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- Inverse of 3 by 3 Matrix

## Introduction

Solving the Inverse of 3 by 3 Matrix may seem tough for some students and hence they might need some Inverse of 3 by 3 Matrix – Solved Examples that will help them understand the concepts and solve the questions well. And for the same reason, we have provided the Inverse of 3 by 3 Matrix sample examples along with some great exercises that you can solve by yourself via Vedantu which will not only help you understand the concepts well but will also prepare you for the exams. Inverse operations are something that is commonly used in algebra to simply what might otherwise be a very complicated question.

## Tips to solve the Inverse of 3 by 3 Matrix

Even though it may seem hard, the best method to overcome it is by solving the question over and over again by using a sample problem.

Write down all the steps and check the answer once again after solving it on your own.

The inverse of a 3 by 3 matrix is a bit complicated task but can be estimated by following the steps given below. A 3 by 3 matrix includes 3 rows and 3 columns. Elements of the matrix are the numbers that form the matrix. A single matrix is one whose determinant is not equivalent to zero. For each x x x square matrix, there exists an inverse of each matrix. The inverse of matrix x * x is represented by X. The inverse of a matrix cannot be easily calculated using a calculator and shortcut method.

\[ XX^{-1} = X^{-1} X = I_{2}\]

In the above property \[I_{2}\] indicates x * x matrix. For example, let us take 2 * 2 matrix as

\[\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}\]

Any x * x square matrix X, which has zero determinant always includes an inverse \[X^{-1}\] .

It is appropriate for almost all the square matrices and is given by \[ XX^{-1} = X^{-1} X = I_{2}\]

## How do you Find the Inverse of the 3 by 3 Matrix?

Here, you can see the inverse of 3 by 3 matrix steps to find the inverse of 3 by 3 matrix online

Estimate the determinant of the given matrix

Find the transpose of the given matrix

Calculate the determinant of the 2 x 2 matrix.

Prepare the matrix of cofactors

At the last, divide each term of the adjugate matrix by the determinant

## Inverse Matrix Formula

The first step is to calculate the determinant of the 3 * 3 matrix and then find its cofactors, minors, and adjoint and then include the results in the below-given inverse matrix formula.

\[A^{-1} = 1/ |A | Adj (A)\]

## Inverse of 3 X 3 Matrix Example

Let us solve the 3 X 3 matrix

\[\begin{bmatrix} a & b &c \\ d & e & f\\ g & h & i\end{bmatrix}\]

Examine the given 3 X 3 matrix

A = \[\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4\\ 5 & 6 & 0\end{bmatrix}\]

Let’s learn the steps to find the inverse of 3 X 3 matrices online

Examine whether the given matrix is invertible

This can be proved if its determinant is non-zero. If the determinant of the given matrix is zero, then there will be no inverse of the given matrix.

det(A) = 1(0-24) – 2(0-20) + 3(0-5)

det(A) = -24 + 40 – 15

det (A) = 1

As the value determinant is 1, we can say that a given matrix has an inverse matrix.

## Finding the Transpose of the Given Matrix

To find the transpose of the given 3 X 3 matrix

Hence,\[A^{-1}\] = \[\begin{bmatrix} 1 & 0 & 5 \\ 2 & 1 & 6\\ 3 & 4 & 0\end{bmatrix}\]

Determining the determinants of the 2 X 2 minor matrices.

Now, we will determine the determinant of each and every 2 X 2 minor matrices.

For first row elements

\[\begin{bmatrix} 1 & 6\\ 4 & 0\end{bmatrix}\] = -24

\[\begin{bmatrix} 12 & 6\\ 3 & 0\end{bmatrix}\] = -18

\[\begin{bmatrix} 2 & 1\\ 3 & 4\end{bmatrix}\] = 5

For second-row elements

\[\begin{bmatrix} 0 & 5\\ 4 & 0\end{bmatrix}\] = -20

\[\begin{bmatrix} 1 & 5\\ 3 & 0\end{bmatrix}\] = -15

\[\begin{bmatrix} 1 & 0\\ 3 & 4\end{bmatrix}\] = 4

For third-row element

\[\begin{bmatrix} 10 & 65\\ 14 & 60\end{bmatrix}\] = -5

\[\begin{bmatrix} 1 & 5\\ 2 & 6\end{bmatrix}\] = -4

\[\begin{bmatrix} 1 & 0\\ 2 & 1\end{bmatrix}\] = 1

Now, the new matrix is

\[\begin{bmatrix} -24 & -18 & 5 \\ -20 & -15 & 4\\ -5 & 14 & 1\end{bmatrix}\]

## Creating the Matrix of Cofactors

To formulate the adjoint or the adjugate matrix, reverse the sign of the alternating terms as shown below:

We get the new matrix as

A = \[\begin{bmatrix} -24 & -18 & 5 \\ -20 & -15 & 4\\ -5 & 14 & 1\end{bmatrix}\]

Adj (A) = the new matrix is

A = \[\begin{bmatrix} -24 & -18 & 5 \\ -20 & -15 & 4\\ -5 & 14 & 1\end{bmatrix}\] X \[\begin{bmatrix} + & - & + \\ - & + & +\\ + & - & +\end{bmatrix}\]

Adj(A) = \[\begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4\\ -5 & 4 & 1\end{bmatrix}\]

Finding the inverse of 3 x 3 matrix

Now, if we will substitute the value of det (A) and the adj (A) in the formula:

\[A^{-1}\] = (1/det (A)) Adj (A)

\[A^{-1}\] = (1/1) = \[\begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4\\ -5 & 4 & 1\end{bmatrix}\]

Hence, the inverse of the given matrix is

## Solved Examples

1. Find the Inverse of the Following Matrix

1 \[\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 1\\ 1 & -1 & 2\end{bmatrix}\]

Solution:

\[ -2\begin{bmatrix} 1 & 1\\ -1 & 2\end{bmatrix} -1\begin{bmatrix} 1 & 1\\ 1 & 2\end{bmatrix} +1\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix} \]

|A | = 2 (2-(-1)) -1 (2-1) +1(-1-1)

= 2 (2+1) -1 (1) +1 (-2)

= 2 (3) -1-2

|A | = 3

As A is a non-singular matrix. \[A^{-1}\] exists

Minor and cofactors of row 1

Minor of 2

\[\begin{bmatrix} 1 & 1\\ -1 & 2\end{bmatrix}\]

Cofactor of 2 = + (3)

\[\begin{bmatrix} 1 & 1\\ 1 & 2\end{bmatrix}\]

Cofactor of 1 = (-1)

\[\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}\]

Cofactor of 1 = + (-2)

Minor and cofactors of row 2

\[\begin{bmatrix} 1 & 1\\ 1 & -2\end{bmatrix}\]

Cofactor of 1 = - (-3)

\[\begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix}\]

\[\begin{bmatrix} 2 & 1\\ 1 & -1\end{bmatrix}\]

Minor and Cofactors of Row 3

\[\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}\]

Cofactor of 1 = + (0)

Minor of -1

\[\begin{bmatrix} 2 & 1\\ 1 & 1\end{bmatrix}\]

Cofactor of -1 = - (1)

Cofactor of 2 = + (1)

Cofactor Matrix

\[\begin{bmatrix} 3 & -1 & -2 \\ -3 & 3 & 3\\ 0 & -1 & 1\end{bmatrix}\]

Adjoint Matrix

\[\begin{bmatrix} 3 & -3 & 0 \\ -1 & 3 & -1\\ -2 & 3 & 1\end{bmatrix}\]

Hence, the inverse of the given matrix is

\[A^{-1}\] = \[\frac{1}{3}\] \[\begin{bmatrix} -3 & 3 & 0 \\ 1 & -3 & 1\\ 2 & -3 & -1\end{bmatrix}\]

\[\begin{bmatrix} 6 & 2 & 3 \\ 3 & 1 & 1\\ 10 & 3 & 4\end{bmatrix}\]

\[2\begin{bmatrix} 1& 1\\ -1& 2 \\\end{bmatrix}\] \[-1\begin{bmatrix} 1 & 1\\ 1 & 2\end{bmatrix} + 1 \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\]

As, A is a non-singular matrix. \[A^{-1}\] exists

|A| = 6 (4-3) - 2 (12-10) + 3 (9-10)

= 6 (1) - 2 (2) + 3 (-1)

= 6 - 4 - 3

|A| = -1 ≠ 0

Minor and Cofactors of Row 1

\[\begin{bmatrix} 1 & 1\\ 3 & 4\end{bmatrix}\]

Cofactor of 6 = + (1)

\[\begin{bmatrix} 3 & 1\\ 10 & 4\end{bmatrix}\]

Cofactor of 2 = (-2)

\[\begin{bmatrix} 3 & 1\\ 10 & 3\end{bmatrix}\]

Cofactor of 3 = + (-1)

Minor and Cofactors of Row 2

Minor of 3

\[\begin{bmatrix} 2 & 3\\ 3 & 4\end{bmatrix}\]

Cofactor of 3 = - (-1)

\[\begin{bmatrix} 6 & 3\\ 10 & 4\end{bmatrix}\]

Cofactor of 1 = + (-6)

\[\begin{bmatrix} 6 & 2\\ 10 & 3\end{bmatrix}\]

Cofactor of 1 = - (-2)

Minor of 10

\[\begin{bmatrix} 2 & 3\\ 1 & 1\end{bmatrix}\]

Cofactor of 1 = + (-1)

\[\begin{bmatrix} 6 & 3\\ 3 & 1\end{bmatrix}\]

Cofactor of 3 = - (-3)

\[\begin{bmatrix} 6 & 2\\ 3 & 1\end{bmatrix}\]

Cofactor of 4 = + (0)

\[\begin{bmatrix} 1 & -2 & -1 \\ -1 & -6 & 2\\ -1 & 3 & 0\end{bmatrix}\]

Adjoint of Matrix

= \[\begin{bmatrix} 1 & 1 & -1 \\ -2 & -6 & 3\\ -1 & 2 & 0\end{bmatrix}\]

\[A^{-1}\] = 1/1 = \[\begin{bmatrix} -1 & -1 & 1 \\ 2 & 6 & -3\\ 1 & -2 & 0\end{bmatrix}\]

## FAQs on Inverse of 3 by 3 Matrix

1. What is the purpose of the Inverse of 3 by 3 Matrix in Algebra?

The inverse of the 3 by 3 Matrix is highly beneficial in Algebraic problems that are highly complicated and needs several steps to be solved. For example, if there is a problem that needs you to divide by a fraction then you can more easily multiply by its reciprocal instead of using the fraction to divide the number. This is called the inverse operation. Similarly, since there is no presence of division that needs to be done in a matrix all you need to do is use the inverse operations and multiply with the help of it. Calculating the Inverse of 3 by 3 Matrix is a very hard task but you can understand and practice them for a better understanding.

2. How to understand the Inverse of 3 by 3 Matrix well with just solved examples?

You don't need to understand the concept involved in the Inverse of 3 by 3 Matrix by just solving the solved examples. Being a student you might have access to the textbook which consists of problems and their related theory. First, make sure you refer to the textbook and check what is provided in it to understand the basic concepts and what the numbers and brackets involved in the Inverse of 3 by 3 Matrix indicate. If you do not have access to the textbook then you can refer to the Vedantu NCERT textbooks provided. Once you understand the concept, then move on to the solved examples that are present in the textbook.

3. Does Vedantu only provide Inverse of 3 by 3 Matrix solved examples or does it also provide other important study materials?

Yes, Vedantu does provide a lot of other study materials other than Inverse of 3 by 3 Matrix solved examples and hence students who do not have access to other extra study materials do not need to worry as they can access all the facilities for free of cost on Vedantu. Students can also get access to the Vedantu sample papers that may contain examples that are similar to Inverse of 3 by 3 Matrix and are deemed to be important for exams.

4. Is it possible to use a calculator to solve questions that are related to the Inverse of 3 by 3 Matrix?

Yes it is possible to use a calculator to solve the problems related to Inverse of 3 by 3 Matrix and the steps can be given as follows

First, select a calculator that consists of matrix capabilities

enter the matrix onto your calculator

Now select the edit submenu given

Select a name for the matrix you have entered

enter the dimension of the matrix entered

enter each of the elements of the matrix

quit the matrix function and use the inverse key to find the inverse matrix from the original matrix

convert the inverse matrix to get the answer to the question.

While it is easy to find the answers using a calculator, not using the calculator is highly recommended so that you understand the concept well.

5. How to get easy shortcuts to solve the Inverse of 3 by 3 Matrix sums that are provided in the Algebra textbook?

Problems based on the Inverse of 3 by 3 Matrix might seem like they are too hard to understand and you need a shortcut so that you can solve them in exams. However, it is seen that there are no shortcuts when it comes to solving Math problems. You not only need to understand the concept but you will also need to understand the steps that are involved while solving the question. Even though it may take some time and practice it is definitely worth all the marks that you score once you solve them in the exam.

## Inverse Matrix Questions

Inverse matrix questions and solutions are given here to help students learn how to find the inverse of different matrices using different formulas and techniques. As we know, matrices are one of the most scoring concepts for students. Finding the inverse matrix is simple for 2×2 matrices. However, we can easily find the inverse matrix for 3×3 and 4×4 matrices using some simple rules. In this article, you will learn how to find the inverse of a given matrix using a suitable method.

What is an inverse matrix?

Suppose A is a non-singular square matrix of order n×n, and there is a matrix B of the same order, such that AB = BA = I, then B is called the inverse matrix of A, and I is the identity matrix.

Inverse matrix formula for 2×2 matrix

|A| = ad – bc

Therefore, A -1 = adjA/|A|

That means,

Inverse matrix formula for 3×3 or n×n matrix

Step 1: Find the determinant of the given matrix, say A.

Step 2: Find the cofactor matrix C ij = (-1) i+j det (M ij ), where M ij is the (i,j)th minor matrix after removing the ith row and the jth column.

Step 3: Find the transpose of the cofactor matrix to get the adj A.

Step 4: A -1 = adj A/det(A)

Learn more about the inverse matrix here.

## Inverse Matrix Questions and Answers

1. Find the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 3 \\1 & 2 \\\end{bmatrix}\end{array} \) .

Let us find the determinant of A.

Here, |A| ≠ 0, so the inverse of A exists.

Now, A -1 = adjA/|A|

2. What is the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}1 & 2 \\-3 & 0 \\\end{bmatrix}\end{array} \) ?

Let us calculate the determinant of A.

As we know, A -1 = adjA/|A|

3. If \(\begin{array}{l}A=\begin{bmatrix}cos\ \theta & sin\ \theta \\-sin\ \theta & cos\ \theta \\\end{bmatrix}\end{array} \) and A -1 = A T , find the value of θ.

Also, given that,

⇒ AA -1 = AA T

From the above,

cos 2 θ + sin 2 θ = 1

This is one of the trigonometric identities and is true for all real values of θ.

4. Calculate the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 4 & -6 \\7 & 3 & 5 \\1 & -2 & 4 \\\end{bmatrix}\end{array} \) .

First, find the determinant of matrix A.

= 2(12 + 10) – 4(28 – 5) – 6(-14 – 3)

= 2(22) – 4(23) – 6(-17)

= 44 – 92 + 102

Thus, the inverse matrix exists.

5. If \(\begin{array}{l}A=\begin{bmatrix}2 & 1 \\7 & 2 \\\end{bmatrix}\end{array} \) , show that (A -1 ) -1 = A.

Here, matrix A is non-singular.

Let A -1 = B

= (-⅔)(-⅔) – (⅓)(7/3)

= (4/9) – (7/9)

= (4 – 7)/9

That means B -1 = (A -1 ) -1 = A

6. Find x, y, z if \(\begin{array}{l}A=\begin{bmatrix}0 & 2y & z \\x & y & -z \\x & -y & z \\\end{bmatrix}\end{array} \) satisfies A T = A -1 .

⇒ AA T = AA -1

⇒ AA T = I {since A -1 A = AA -1 = I}

By performing multiplication on the LHS, we get:

By equating the corresponding elements, we have:

4y 2 + z 2 = 1 …(1)

x 2 + y 2 + z 2 = 1 …(2)

2y 2 – z 2 = 0 …(3)

Adding equations (1) and (3), we get:

4y 2 + z 2 + 2y 2 – z 2 = 1 + 0

⇒ y = ±1/√6

Substituting the value of y in equation (3), we get:

z 2 = 2(1/6)

⇒ z = ±1/√3

Substituting the values of y and z in equation (2), we get:

x 2 = 1 – y 2 – z 2

x 2 = 1 – (1/6) – (1/3)

x 2 = (6 – 1 – 2)/6

⇒ x = ±1/√2

Therefore, x = ±1/√2, y = ±1/√6 and z = ±1/√3.

7. Find the value of x for which the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 0 & 10 \\ 0 & x+7 & -3 \\0 & 4 & x \\\end{bmatrix}\end{array} \) is invertible.

Let us find the determinant of the given matrix.

= 2[(x + 7)x – (-3)(4)] – 0 + 10(0 – 0)

= 2(x 2 + 7x + 12)

We know that a matrix is invertible if and only if its determinant is not equal to 0.

Let |A| = 0

2(x 2 + 7x + 12) = 0

⇒ x 2 + 7x + 12 = 0

⇒ x 2 + 3x + 4x + 12 = 0

⇒ x(x + 3) + 4(x + 3) = 0

⇒ (x + 3)(x + 4) = 0

⇒ x + 3 = 0, x + 4 = 0

⇒ x = -3, x = -4

Thus, for x = -3 and -4, the given matrix is invertible.

8. Find the inverse of \(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 3 \\2 & 4 & 5 \\3 & 5 & 6 \\\end{bmatrix}\end{array} \) using row operations.

Let us write the augmented matrix [A | I ] such that I is a square matrix of the order same as A.

R 3 → R 3 – 3R 1

R 2 → R 2 – 2R 1

Now, interchange R 2 and R 3 .

R 2 → (-1).R 2 and R 3 → (-1).R 3

R 2 → R 2 – 3R 3

R 1 → R 1 – 2R 2

R 1 → R 1 – 3R 3

This is of the form [ I | B].

Here, B is the inverse of A.

9. Determine the formula for the inverse of matrix \(\begin{array}{l}A=\begin{bmatrix}p & 0 & 0 & 0 \\0 & q & 0 & 0 \\0 & 0 & r & 0 \\0 & 0 & 0 & s \\\end{bmatrix}\end{array} \) , where p, q, r, s ≠ 0.

Let us write the augmented matrix [A | I ].

R 1 → (1/p) R 1

R 2 → (1/q) R 2

R 3 → (1/r) R 3

R 4 → (1/s) R 4

Hence, the inverse of A is:

10. If A is 3 × 3 invertible matrix, then show that for any scalar k (non-zero), kA is invertible and (kA) –1 = (1/k)A –1 .

= [k (1/k)] (A A -1 )

= 1. (AA -1 )

= I {since AA -1 = A -1 A = I}

That means kA is the inverse of (1/k)A -1 .

Therefore, (kA) -1 = (1/k) A -1

## Practice Questions on Inverse Matrix

- Find the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}3 & 5 \\-2 & 4 \\\end{bmatrix}\end{array} \) .
- If \(\begin{array}{l}A=\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\\\end{bmatrix}\end{array} \) , find A -1 exists.
- Using elementary row operations, find the inverse of the matrix \(\begin{array}{l}\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\\\end{bmatrix}\end{array} \) .
- Calculate the inverse matrix for \(\begin{array}{l}B=\begin{bmatrix}4+3i & -i \\i &4-3i \\\end{bmatrix}\end{array} \) .
- What is the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}1 & 0 & 1 & 2 \\-1 & 1 & 2 & 0 \\-2 & 0 & 1 & 2 \\0 & 0 & 0 & 1 \\\end{bmatrix}\end{array} \) ?

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Problems in Mathematics

- Inverse Matrices
- An $n\times n$ matrix $A$ is said to be invertible if there exists an $n\times n$ matrix $B$ such that $AB=BA=I$. Such a matrix $B$ is unique and called the inverse matrix of $A$, denoted by $A^{-1}$

Let $A, B$ be $n\times n$ matrices.

- $A$ is invertible if and only if $\rref([ A \mid I_n])=[ I_n \mid A’]$ for some $n\times n$ matrix $A’$. In this case, $A’=A^{-1}$.
- $A$ is invertible if and only if $A$ is nonsingular.
- If $A, B$ are invertible, then $(AB)^{-1}=B^{-1}A^{-1}$
- A $2\times 2$ matrix $A=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$ is invertible if and only if the determinant $\det(A)=ad-bc \neq 0$. If $A$ is invertible, then the inverse matrix is given by $A^{-1}=\frac{1}{\det(A)}\begin{bmatrix} d & -b\\ -c& a \end{bmatrix}$.
- If $A$ is invertible, then $A^{\trans}$ is invertible and $(A^{\trans})^{-1}=(A^{-1})^{\trans}$.
- For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix. (a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$ (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.
- Let A be the matrix \[\begin{bmatrix} 1 & -1 & 0 \\ 0 &1 &-1 \\ 0 & 0 & 1 \end{bmatrix}.\] Is the matrix $A$ invertible? If not, then explain why it isn’t invertible. If so, then find the inverse. ( The Ohio State University )
- Find the inverse matrix of \[A=\begin{bmatrix} 1 & 1 & 2 \\ 0 &0 &1 \\ 1 & 0 & 1 \end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason. ( The Ohio State University )
- Let $A$ be the following $3\times 3$ upper triangular matrix. \[A=\begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 \end{bmatrix},\] where $x, y, z$ are some real numbers. Determine whether the matrix $A$ is invertible or not. If it is invertible, then find the inverse matrix $A^{-1}$.
- For which choice(s) of the constant $k$ is the following matrix invertible? \[A=\begin{bmatrix} 1 & 1 & 1 \\ 1 &2 &k \\ 1 & 4 & k^2 \end{bmatrix}.\] ( Johns Hopkins University )
- Suppose that $M, P$ are two $n \times n$ non-singular matrix. Prove that there is a matrix $N$ such that $MN = P$.
- Let $A$ be an $n \times n$ matrix satisfying $A^2+c_1A+c_0I=O$, where $c_0, c_1$ are scalars, $I$ is the $n\times n$ identity matrix, and $O$ is the $n\times n$ zero matrix. Prove that if $c_0\neq 0$, then the matrix $A$ is invertible (nonsingular). How about the converse? Namely, is it true that if $c_0=0$, then the matrix $A$ is not invertible?
- A square matrix $A$ is called idempotent if $A^2=A$. Show that a square invertible idempotent matrix is the identity matrix.
- Find the inverse matrix of $A=\begin{bmatrix} 1 & 0 & 1 \\ 1 &0 &0 \\ 2 & 1 & 1 \end{bmatrix}$ if it exists. If you think there is no inverse matrix of $A$, then give a reason. See (a)
- Find a nonsingular $2\times 2$ matrix $A$ such that $A^3=A^2B-3A^2$, where $B=\begin{bmatrix} 4 & 1\\ 2& 6 \end{bmatrix}$. Verify that the matrix $A$ you obtained is actually a nonsingular matrix. See (b)
- Determine whether there exists a nonsingular matrix $A$ if $A^2=AB+2A$, where $B$ is the following matrix. If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$. (a) \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 1 & 2 & -2 \end{bmatrix}\] (b) \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.\]
- Determine whether there exists a nonsingular matrix $A$ if $A^4=ABA^2+2A^3$, where $B$ is the following matrix. $B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.$ If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$. ( The Ohio State University, Linear Algebra Final Exam Problem )
- Let $A, B, C$ be the following $3\times 3$ matrices. \[A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \\ 7 & 8 & 9 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 & 1 \\ 0 &3 &0 \\ 1 & 0 & 5 \end{bmatrix}, C=\begin{bmatrix} -1 & 0\ & 1 \\ 0 &5 &6 \\ 3 & 0 & 1 \end{bmatrix}.\] Then compute and simplify the following expression. \[(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}.\] ( The Ohio State University )
- Let $A$ be the coefficient matrix of the system of linear equations \begin{align*} -x_1-2x_2&=1\\ 2x_1+3x_2&=-1. \end{align*} (a) Solve the system by finding the inverse matrix $A^{-1}$. (b) Let $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ be the solution of the system obtained in part (a). Calculate and simplify \[A^{2017}\mathbf{x}.\] ( The Ohio State University )
- (a) Let $A$ be a $6\times 6$ matrix and suppose that $A$ can be written as $A=BC$, where $B$ is a $6\times 5$ matrix and $C$ is a $5\times 6$ matrix. Prove that the matrix $A$ cannot be invertible. (b) Let $A$ be a $2\times 2$ matrix and suppose that $A$ can be written as $A=BC$, where $B$ is a $ 2\times 3$ matrix and $C$ is a $3\times 2$ matrix. Can the matrix $A$ be invertible?
- For a real number $a$, consider $2\times 2$ matrices $A, P, Q$ satisfying the following five conditions. (1) $A=aP+(a+1)Q$, (2) $P^2=P$, (3) $Q^2=Q$, (4) $PQ=O$, (5) $QP=O$, where $O$ is the $2\times 2$ zero matrix. Then do the following problems. (a) Prove that $(P+Q)A=A$. (b) Suppose $a$ is a positive real number and let $A=\begin{bmatrix} a & 0\\ 1& a+1 \end{bmatrix}$. Then find all matrices $P, Q$ satisfying conditions (1)-(5). (c) Let $n$ be an integer greater than $1$. For any integer $k$, $2\leq k \leq n$, we define the matrix $A_k=\begin{bmatrix} k & 0\\ 1& k+1 \end{bmatrix}$. Then calculate and simplify the matrix product $A_nA_{n-1}A_{n-2}\cdots A_2$. ( Tokyo University Entrance Exam 2007 )
- Let $\mathbf{v}$ be a nonzero vector in $\R^n$. Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by $A=I-a\mathbf{v}\mathbf{v}^{\trans}$, where $I$ is the $n\times n$ identity matrix. Prove that $A$ is a symmetric matrix and $AA=I$. Conclude that the inverse matrix is $A^{-1}=A$.
- Consider the system of linear equations \begin{align*} x_1&= 2, \\ -2x_1 + x_2 &= 3, \\ 5x_1-4x_2 +x_3 &= 2 \end{align*} (a) Find the coefficient matrix and its inverse matrix. (b) Using the inverse matrix, solve the system of linear equations. ( The Ohio State University )
- Consider the following system of linear equations \begin{align*} 2x+3y+z&=-1\\ 3x+3y+z&=1\\ 2x+4y+z&=-2. \end{align*} (a) Find the coefficient matrix $A$ for this system. (b) Find the inverse matrix of the coefficient matrix found in (a) (c) Solve the system using the inverse matrix $A^{-1}$.
- Consider the matrix \[A=\begin{bmatrix} 1 & 2 & 1 \\ 2 &5 &4 \\ 1 & 1 & 0 \end{bmatrix}.\] (a) Calculate the inverse matrix $A^{-1}$. If you think the matrix $A$ is not invertible, then explain why. (b) Are the vectors \[ \mathbf{A}_1=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \mathbf{A}_2=\begin{bmatrix} 2 \\ 5 \\ 1 \end{bmatrix}, \text{ and } \mathbf{A}_3=\begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}\] linearly independent? (c) Write the vector $\mathbf{b}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ as a linear combination of $\mathbf{A}_1$, $\mathbf{A}_2$, and $\mathbf{A}_3$. ( The Ohio State University, Linear Algebra Exam )
- A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. See (a)
- Suppose that a real matrix $A$ maps each of the following vectors \[\mathbf{x}_1=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \mathbf{x}_2=\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \mathbf{x}_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \] into the vectors \[\mathbf{y}_1=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \mathbf{y}_2=\begin{bmatrix} -1 \\ 0 \\ 3 \end{bmatrix}, \mathbf{y}_3=\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix},\] respectively. That is, $A\mathbf{x}_i=\mathbf{y}_i$ for $i=1,2,3$. Find the matrix $A$. ( Kyoto University Exam )
- Let $A$ and $B$ are $n \times n$ matrices with real entries. Assume that $A+B$ is invertible. Then show that \[A(A+B)^{-1}B=B(A+B)^{-1}A.\] ( University of California, Berkeley Qualifying Exam )
- Let $A$ be an $n\times n$ invertible matrix. Prove that the inverse matrix of $A$ is uniques.
- Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies \[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\] Define the matrix \[A=I+\mathbf{u}\mathbf{v}^{\trans}.\] Prove that $A$ is invertible and the inverse matrix is given by the formula \[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans},\] where \[a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}.\] This formula is called the Sherman-Woodberry formula .
- Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix. Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\] Using the formula, calculate the inverse matrix of $\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}$.
- Let $A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$. Show that (a) $A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix}$ for any $n \in \N$. (b) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix. Show that $B^n=S^{-1}A^n S$ for any $n \in \N$
- Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$. Namely, show that $(A^{\trans})^{-1}=(A^{-1})^{\trans}$.
- A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. (a) If $A$ is a nilpotent $n \times n$ matrix and $B$ is an $n\times n$ matrix such that $AB=BA$. Show that the product $AB$ is nilpotent. (b) Let $P$ be an invertible $n \times n$ matrix and let $N$ be a nilpotent $n\times n$ matrix. Is the product $PN$ nilpotent? If so, prove it. If not, give a counterexample.
- (a) Show that if $A$ is invertible, then $A$ is nonsingular. (b) Let $A, B, C$ be $n\times n$ matrices such that $AB=C$. Prove that if either $A$ or $B$ is singular, then so is $C$. (c) Show that if $A$ is nonsingular, then $A$ is invertible.
- A square matrix $A$ is called nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$. Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample.
- Let $A$ be an $n\times n$ matrix. The $(i, j)$ cofactor $C_{ij}$ of $A$ is defined to be $C_{ij}=(-1)^{ij}\det(M_{ij})$, where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column. Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$. The matrix $\Adj(A)$ is called the adjoint matrix of $A$. When $A$ is invertible, then its inverse can be obtained by the formula \[A^{-1}=\frac{1}{\det(A)}\Adj(A).\] For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula. (a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$.
- Let $A$ be a real symmetric matrix whose diagonal entries are all positive real numbers. Is it true that the all of the diagonal entries of the inverse matrix $A^{-1}$ are also positive? If so, prove it. Otherwise, give a counterexample.
- Let $A$ be an $n\times n$ nonsingular matrix with integer entries. Prove that the inverse matrix $A^{-1}$ contains only integer entries if and only if $\det(A)=\pm 1$.
- Introduction to Matrices
- Elementary Row Operations
- Gaussian-Jordan Elimination
- Solutions of Systems of Linear Equations
- Linear Combination and Linear Independence
- Nonsingular Matrices
- Subspaces in $\R^n$
- Bases and Dimension of Subspaces in $\R^n$
- General Vector Spaces
- Subspaces in General Vector Spaces
- Linearly Independency of General Vectors
- Bases and Coordinate Vectors
- Dimensions of General Vector Spaces
- Linear Transformation from $\R^n$ to $\R^m$
- Linear Transformation Between Vector Spaces
- Orthogonal Bases
- Determinants of Matrices
- Computations of Determinants
- Introduction to Eigenvalues and Eigenvectors
- Eigenvectors and Eigenspaces
- Diagonalization of Matrices
- The Cayley-Hamilton Theorem
- Dot Products and Length of Vectors
- Eigenvalues and Eigenvectors of Linear Transformations
- Jordan Canonical Form

## Module 14: Solve Systems With Matrices

Solving a system using an inverse, learning outcomes.

- Solve a 2×2 system using an inverse.
- Solve a 3×3 systems using an inverse.
- Solve a system with a calculator.

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: [latex]X[/latex] is the matrix representing the variables of the system, and [latex]B[/latex] is the matrix representing the constants. Using matrix multiplication , we may define a system of equations with the same number of equations as variables as [latex]AX=B[/latex]

To solve a system of linear equations using an inverse matrix , let [latex]A[/latex] be the coefficient matrix , let [latex]X[/latex] be the variable matrix, and let [latex]B[/latex] be the constant matrix. Thus, we want to solve a system [latex]AX=B[/latex]. For example, look at the following system of equations.

[latex]\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}[/latex]

From this system, the coefficient matrix is

[latex]A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right][/latex]

The variable matrix is

[latex]X=\left[\begin{array}{c}x\\ y\end{array}\right][/latex]

And the constant matrix is

[latex]B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][/latex]

Then [latex]AX=B[/latex] looks like

[latex]\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][/latex]

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex]\left({2}^{-1}\right)2=\left(\frac{1}{2}\right)2=1[/latex]. To solve a single linear equation [latex]ax=b[/latex] for [latex]x[/latex], we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of [latex]a[/latex]. Thus,

[latex]\begin{array}{c}\text{ }ax=b\\ \text{ }\left(\frac{1}{a}\right)ax=\left(\frac{1}{a}\right)b\\ \left({a}^{-1}\text{ }\right)ax=\left({a}^{-1}\right)b\\ \left[\left({a}^{-1}\right)a\right]x=\left({a}^{-1}\right)b\\ \text{ }1x=\left({a}^{-1}\right)b\\ \text{ }x=\left({a}^{-1}\right)b\end{array}[/latex]

The only difference between solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.

We will investigate this idea in detail, but it is helpful to begin with a [latex]2\times 2[/latex] system and then move on to a [latex]3\times 3[/latex] system.

## A General Note: Solving a System of Equations Using the Inverse of a Matrix

Given a system of equations, write the coefficient matrix [latex]A[/latex], the variable matrix [latex]X[/latex], and the constant matrix [latex]B[/latex]. Then [latex]AX=B[/latex]. Multiply both sides by the inverse of [latex]A[/latex] to obtain the solution.

[latex]\begin{array}{r}\hfill \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \hfill \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \hfill IX=\left({A}^{-1}\right)B\\ \hfill X=\left({A}^{-1}\right)B\end{array}[/latex]

If the coefficient matrix does not have an inverse, does that mean the system has no solution?

No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution or be dependent and have infinitely many solutions.

## Example: Solving a 2 × 2 System Using the Inverse of a Matrix

Solve the given system of equations using the inverse of a matrix.

[latex]\begin{array}{r}\hfill 3x+8y=5\\ \hfill 4x+11y=7\end{array}[/latex]

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

[latex]A=\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}5\\ 7\end{array}\right][/latex]

[latex]\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 7\end{array}\right][/latex]

First, we need to calculate [latex]{A}^{-1}[/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

[latex]\begin{array}{l}{A}^{-1} & =\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\hfill \\ & =\frac{1}{3\left(11\right)-8\left(4\right)}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\hfill \\ & =\frac{1}{1}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\hfill \end{array}[/latex]

[latex]{A}^{-1}=\left[\begin{array}{cc}11& -8\\ -4& \text{ }\text{ }3\end{array}\right][/latex]

Now we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[/latex].

[latex]\begin{array}{l}\left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\hfill \\ \left[\begin{array}{rr}\hfill 11& \hfill -8\\ \hfill -4& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{rr}\hfill 11& \hfill -8\\ \hfill -4& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{c}5\\ 7\end{array}\right]\hfill \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\hfill 11\left(5\right)+\left(-8\right)7\\ \hfill -4\left(5\right)+3\left(7\right)\end{array}\right]\hfill \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\hfill -1\\ \hfill 1\end{array}\right]\hfill \end{array}[/latex]

The solution is [latex]\left(-1,1\right)[/latex].

Can we solve for [latex]X[/latex] by finding the product [latex]B{A}^{-1}?[/latex]

No, recall that matrix multiplication is not commutative, so [latex]{A}^{-1}B\ne B{A}^{-1}[/latex]. Consider our steps for solving the matrix equation.

Notice in the first step we multiplied both sides of the equation by [latex]{A}^{-1}[/latex], but the [latex]{A}^{-1}[/latex] was to the left of [latex]A[/latex] on the left side and to the left of [latex]B[/latex] on the right side. Because matrix multiplication is not commutative, order matters.

## Example: Solving a 3 × 3 System Using the Inverse of a Matrix

Solve the following system using the inverse of a matrix.

[latex]\begin{array}{r}\hfill 5x+15y+56z=35\\ \hfill -4x - 11y - 41z=-26\\ \hfill -x - 3y - 11z=-7\end{array}[/latex]

Write the equation [latex]AX=B[/latex].

[latex]\left[\begin{array}{ccc}5& 15& 56\\ -4& -11& -41\\ -1& -3& -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right][/latex]

First, we will find the inverse of [latex]A[/latex] by augmenting with the identity.

[latex]\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

Multiply row 1 by [latex]\frac{1}{5}[/latex].

[latex]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ -4& -11& -41\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

Multiply row 1 by 4 and add to row 2.

[latex]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ 0& 0& 1\end{array}\right][/latex]

Add row 1 to row 3.

[latex]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right][/latex]

Multiply row 2 by −3 and add to row 1.

[latex]\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right][/latex]

Multiply row 3 by 5.

[latex]\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right][/latex]

Multiply row 3 by [latex]\frac{1}{5}[/latex] and add to row 1.

[latex]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right][/latex]

Multiply row 3 by [latex]-\frac{19}{5}[/latex] and add to row 2.

[latex]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right][/latex]

[latex]{A}^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right][/latex]

Multiply both sides of the equation by [latex]{A}^{-1}[/latex]. We want

[latex]{A}^{-1}AX={A}^{-1}B:[/latex]

[latex]\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill 5& \hfill 15& \hfill 56\\ \hfill -4& \hfill -11& \hfill -41\\ \hfill -1& \hfill -3& \hfill -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{rrr}\hfill -2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 1& \hfill -19\\ \hfill 1& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{r}\hfill 35\\ \hfill -26\\ \hfill -7\end{array}\right][/latex]

[latex]{A}^{-1}B=\left[\begin{array}{r}\hfill -70+78 - 7\\ \hfill -105 - 26+133\\ \hfill 35+0 - 35\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right][/latex]

The solution is [latex]\left(1,2,0\right)[/latex].

Solve the system using the inverse of the coefficient matrix.

[latex]\begin{array}{l}\text{ }2x - 17y+11z=0\hfill \\ \text{ }-x+11y - 7z=8\hfill \\ \text{ }3y - 2z=-2\hfill \end{array}[/latex]

[latex]X=\left[\begin{array}{c}4\\ 38\\ 58\end{array}\right][/latex]

## How To: Given a system of equations, solve with matrix inverses using a calculator

- Save the coefficient matrix and the constant matrix as matrix variables [latex]\left[A\right][/latex] and [latex]\left[B\right][/latex].
- Enter the multiplication into the calculator, calling up each matrix variable as needed.
- If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.

## Example: Using a Calculator to Solve a System of Equations with Matrix Inverses

Solve the system of equations with matrix inverses using a calculator

[latex]\begin{array}{l}2x+3y+z=32\hfill \\ 3x+3y+z=-27\hfill \\ 2x+4y+z=-2\hfill \end{array}[/latex]

On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [latex]\left[A\right][/latex], and enter the constant matrix as the matrix variable [latex]\left[B\right][/latex].

[latex]\left[A\right]=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right],\text{ }\left[B\right]=\left[\begin{array}{c}32\\ -27\\ -2\end{array}\right][/latex]

On the home screen of the calculator, type in the multiplication to solve for [latex]X[/latex], calling up each matrix variable as needed.

[latex]{\left[A\right]}^{-1}\times \left[B\right][/latex]

Evaluate the expression.

[latex]\left[\begin{array}{c}-59\\ -34\\ 252\end{array}\right][/latex]

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## Inverse of Matrix

The inverse of Matrix for a matrix A is denoted by A -1 . The inverse of a 2 × 2 matrix can be calculated using a simple formula. Further, to find the inverse of a matrix of order 3 or higher, we need to know about the determinant and adjoint of the matrix. The inverse of a matrix is another matrix, which by multiplying with the given matrix gives the identity matrix.

The inverse of matrix is used of find the solution of linear equations through the matrix inversion method. Here, let us learn about the formula, methods, and terms related to the inverse of matrix.

## What is Inverse of Matrix?

The inverse of matrix is a matrix, which on multiplication with the given matrix gives the multiplicative identity . For a square matrix A, its inverse is A -1 , and A · A -1 = A -1 · A = I, where I is the identity matrix. The matrix whose determinant is non-zero and for which the inverse matrix can be calculated is called an invertible matrix. For example, the inverse of A = \(\left[\begin{array}{rr} 1 & -1 \\ \\ 0 & 2 \end{array}\right]\) is \(\left[\begin{array}{cc} 1 & 1 / 2 \\ \\ 0 & 1 / 2 \end{array}\right]\) as

- A · A -1 = \(\left[\begin{array}{rr} 1 & -1 \\ \\ 0 & 2 \end{array}\right]\) \(\left[\begin{array}{cc} 1 & 1 / 2 \\ \\ 0 & 1 / 2 \end{array}\right]\) = \(\left[\begin{array}{cc} 1 & 0 \\ \\ 0 & 1 \end{array}\right]\) = I
- A -1 · A = \(\left[\begin{array}{cc} 1 & 1 / 2 \\ \\ 0 & 1 / 2 \end{array}\right]\) \(\left[\begin{array}{rr} 1 & -1 \\ \\ 0 & 2 \end{array}\right]\) = \(\left[\begin{array}{cc} 1 & 0 \\ \\ 0 & 1 \end{array}\right]\) = I

But how to find the inverse of a matrix? Let us see in the upcoming sections.

## Inverse Matrix Formula

In the case of real numbers , the inverse of any real number a was the number a -1 , such that a times a -1 equals 1. We knew that for a real number, the inverse of the number was the reciprocal of the number, as long as the number wasn't zero. The inverse of a square matrix A, denoted by A -1 , is the matrix so that the product of A and A -1 is the identity matrix . The identity matrix that results will be the same size as matrix A.

The formula to find the inverse of a matrix is: A -1 = 1/|A| · Adj A, where

- |A| is the determinant of A and
- Adj A is the adjoint of A

Since |A| is in the denominator of the above formula, the inverse of a matrix exists only if the determinant of the matrix is a non-zero value. i.e., |A| ≠ 0.

## How to Find Matrix Inverse?

To find the inverse of a square matrix A, we use the following formula: A -1 = adj(A) / |A| ; |A| ≠ 0

- A is a square matrix.
- adj(A) is the adjoint matrix of A.
- |A| is the determinant of A.

☛ Note: For a matrix to have its inverse exists:

- The given matrix should be a square matrix.
- The determinant of the matrix should not be equal to zero.

## Terms Related to Matrix Inverse

The following terms below are helpful for more clear understanding and easy calculation of the inverse of matrix.

Minor: The minor is defined for every element of a matrix . The minor of a particular element is the determinant obtained after eliminating the row and column containing this element. For a matrix A = \(\begin{pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}\), the minor of the element \(a_{11}\) is:

Minor of \(a_{11}\) = \(\left|\begin{matrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{matrix}\right|\)

Cofactor: The cofactor of an element is calculated by multiplying the minor with -1 to the exponent of the sum of the row and column elements in order representation of that element.

Cofactor of \(a_{ij}\) = (-1) i + j × minor of \(a_{ij}\).

Determinant: The determinant of a matrix is the single unique value representation of a matrix. The determinant of the matrix can be calculated with reference to any row or column of the given matrix. The determinant of the matrix is equal to the summation of the product of the elements and its cofactors, of a particular row or column of the matrix.

Singular Matrix: A matrix having a determinant value of zero is referred to as a singular matrix . For a singular matrix A, |A| = 0. The inverse of a singular matrix does not exist.

Non-Singular Matrix: A matrix whose determinant value is not equal to zero is referred to as a non-singular matrix . For a non-singular matrix |A| ≠ 0 and hence its inverse exists.

Adjoint of Matrix: The adjoint of a matrix is the transpose of the cofactor element matrix of the given matrix.

Rules For Row and Column Operations of a Determinant:

The following rules are helpful to perform the row and column operations on determinants .

- The value of the determinant remains unchanged if the rows and columns are interchanged.
- The sign of the determinant changes, if any two rows or (two columns) are interchanged.
- If any two rows or columns of a matrix are equal, then the value of the determinant is zero.
- If every element of a particular row or column is multiplied by a constant , then the value of the determinant also gets multiplied by the constant.
- If the elements of a row or a column are expressed as a sum of elements, then the determinant can be expressed as a sum of determinants.
- If the elements of a row or column are added or subtracted with the corresponding multiples of elements of another row or column, then the value of the determinant remains unchanged.

## Methods to Find Inverse of Matrix

The inverse of a matrix can be found using two methods. The inverse of a matrix can be calculated through elementary operations and through the use of an adjoint of a matrix. The elementary operations on a matrix can be performed through row or column transformations. Also, the inverse of a matrix can be calculated by applying the inverse of matrix formula through the use of the determinant and the adjoint of the matrix. For performing the inverse of the matrix through elementary column operations we use the matrix X and the second matrix B on the right-hand side of the equation.

- Elementary row or column operations
- Inverse of matrix formula (using the adjoint and determinant of matrix)

Let us check each of the methods described below.

Elementary Row Operations

To calculate the inverse of matrix A using elementary row transformations, we first take the augmented matrix [A | I], where I is the identity matrix whose order is the same as A. Then we apply the row operations to convert the left side A into I. Then the matrix gets converted into [I | A -1 ]. For a more detailed process, click here .

Elementary Column Operations

We can apply the column operations as well just like how the process was explained for row operations to find the inverse of matrix.

Inverse of Matrix Formula

The inverse of matrix A can be computed using the inverse of matrix formula, A -1 = (adj A)/(det A). i.e., by dividing the adjoint of a matrix by the determinant of the matrix. The inverse of a matrix can be calculated by following the given steps:

- Step 1: Calculate the minors of all elements of A.
- Step 2: Then compute the cofactors of all elements and write the cofactor matrix by replacing the elements of A by their corresponding cofactors.
- Step 3: Find the adjoint of A (written as adj A) by taking the transpose of the cofactor matrix of A.
- Step 4: Multiply adj A by the reciprocal of the determinant.

For a matrix A, its inverse A -1 = \(\dfrac{1}{|A|}\) · Adj A

If A = \(\begin{pmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}\), then

- |A| = \(a_{11}(-1)^{1 + 1} \left|\begin{matrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{matrix}\right| + a_{12}(-1)^{1 + 2} \left|\begin{matrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{matrix}\right| + a_{13}(-1)^{1 + 3} \left|\begin{matrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{matrix}\right|\)
- Adj A = Transpose of Cofactor Matrix = Transpose of \(\begin{pmatrix} A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{pmatrix}\) =\(\begin{pmatrix} A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{pmatrix}\)

A -1 = \(\dfrac{1}{|A|}.\begin{pmatrix} A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{pmatrix}\)

In this section, we have learned the different methods to calculate the inverse of a matrix. Let us understand it better using a few examples for the different orders of matrices in the " examples " section below.

## Inverse of 2 x 2 Matrix

The inverse of 2 × 2 matrix is easier to calculate in comparison to matrices of higher order. We can calculate the inverse of 2 × 2 matrix using the general steps to calculate the inverse of a matrix. Let us find the inverse of the 2 × 2 matrix given below: A = \(\begin{bmatrix} a & b \\ \\ c & d \end{bmatrix}\) A -1 = (1/|A|) × Adj A = [1/(ad - bc)] × \(\begin{bmatrix} d & -b \\ \\ -c & a \end{bmatrix}\)

Therefore, in order to calculate the inverse of 2 × 2 matrix, we need to first swap the positions of terms a and d and put negative signs for terms b and c, and finally divide it by the determinant of the matrix.

## Inverse of 3 x 3 Matrix

We know that for every non-singular square matrix A, there exists an inverse matrix A -1 , such that A × A -1 = I. Let us take any 3 × 3 square matrix given as,

A = \(\begin{bmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\)

The inverse of 3x3 matrix can be calculated using the inverse matrix formula, A -1 = (1/|A|) × Adj A

We will first check if the given matrix is invertible, i.e., |A| ≠ 0. If the inverse of a matrix exists, we can find the adjoint of the given matrix and divide it by the determinant of the matrix.

A similar method can be followed to find the inverse of any n × n matrix. Let us see if similar steps can be used to calculate the inverse of m × n matrix where m ≠ n.

## Inverse of 2 × 3 Matrix

We know that the first condition for the inverse of a matrix to exist is that the given matrix should be a square matrix. Also, the determinant of this square matrix should be non-zero. This means that the inverse of matrices of the order m × n will not exist where m ≠ n. Therefore, we cannot calculate the inverse of 2 × 3 matrix.

## Inverse of 2 × 1 Matrix

Similar to the inverse of 2 × 3 matrix, the inverse of 2 × 1 matrix also does not exist because the given matrix is not a square matrix.

## Determinant of Inverse Matrix

The determinant of the inverse of an invertible matrix is the inverse of the determinant of the original matrix. i.e., det(A -1 ) = 1 / det(A). Let us check the proof of the above statement.

We know that, det(A • B) = det (A) × det(B)

Also, A × A -1 = I

det(A × A -1 ) = det(I)

or, det(A) × det(A -1 ) = det(I)

Since, det(I) = 1

det(A) × det(A -1 ) = 1

or, det(A -1 ) = 1 / det(A)

Hence, proved.

☛ Related Articles:

- Inverse Matrix Calculator
- Matrix Multiplication Calculator
- Determinant Calculator

Important Points on Inverse of a Matrix:

- The inverse of a square matrix (if exists) is unique.
- If A and B are two invertible matrices of the same order then (AB) -1 = B -1 A -1 .
- The inverse of a square matrix A exists, only if its determinant is a non-zero value, |A| ≠ 0.
- The determinant of matrix inverse is equal to the reciprocal of the determinant of the original matrix.
- The determinant of the product of two matrices is equal to the product of the determinants of the two individual matrices. |AB| = |A|.|B|

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## Inverse of Matrix Examples

Example 1: Find the inverse of matrix A = \(\left(\begin{matrix}-3 & 4\\ \\ 2 & 5 \end{matrix}\right)\).

The given matrix is A = \(\left(\begin{matrix}-3 & 4\\ \\2 & 5 \end{matrix}\right)\).

The formula to calculate the matrix inverse of A = \(\left(\begin{matrix}a&b\\\\c&d\end{matrix}\right)\) is A -1 = \(\dfrac{1}{ad - bc}\left(\begin{matrix}d&-b\\\\-c&a\end{matrix}\right)\).

Using this formula we can calculate A -1 as follows.

A -1 = \(\dfrac{1}{(-3)× 5 - 4 × 2}\left(\begin{matrix}5&-4\\\\-2&-3\end{matrix}\right)\)

= \(\dfrac{1}{-15 - 8}\left(\begin{matrix}5&-4\\\\-2&-3\end{matrix}\right)\)

= \(\dfrac{-1}{23}\left(\begin{matrix}5&-4\\\\-2&-3\end{matrix}\right)\)

Answer: Therefore A -1 = \(\dfrac{-1}{23}\left(\begin{matrix}5&-4\\\\-2&-3\end{matrix}\right)\)

Example 2: Find the matrix inverse of the matrix A = \(\left(\begin{matrix}4 & -2 & 1\\5&0&3\\-1&2 & 6\end{matrix}\right)\).

The given matrix is A = \(\left(\begin{matrix}4 & -2 & 1\\5&0&3\\-1&2 & 6\end{matrix}\right)\)

Step - 1: Let us find the determinant of the given matrix using Row - 1 of the above matrix.

|A| = \(4\left|\begin{matrix}0&3\\2 & 6\end{matrix}\right| -(-2)\left|\begin{matrix}5&3\\-1 & 6\end{matrix}\right|+1\left|\begin{matrix}5&0\\-1& 2\end{matrix}\right|\)

= 4(0 × 6 - 3 × 2) + 2(5 × 6 - (-1) × 3) +1(5 × 2 - 0 × (-1))

= 4(0 - 6) + 2(30 + 3) + 1(10 - 0)

= -24 + 66 + 10

Now, we will determine the adjoint of the matrix A by calculating the cofactors of each element and then taking the transpose of the cofactor matrix.

Adj A = \(\left(\begin{matrix}-6 & 14 & -6\\-33&25&-7\\10&-6 & 10\end{matrix}\right)\)

The inverse of matrix A is given by the formula A -1 = \(\dfrac{1}{|A|}\).Adj A

A -1 = \(\dfrac{1}{52}\).\(\left(\begin{matrix}-6 & 14 & -6\\-33&25&-7\\10&-6 & 10\end{matrix}\right)\)

= \(\left(\begin{matrix}-3/26 & 7/26 & -3/26\\-33/52&25/52&-7/52\\5/26&-3/26 & 5/26\end{matrix}\right)\)

Answer: A -1 = \(\left(\begin{matrix}-3/26 & 7/26 & -3/26\\-33/52&25/52&-7/52\\5/26&-3/26 & 5/26\end{matrix}\right)\)

Example 3: Find the inverse of \(\begin{bmatrix} 4 & 2 \\\\ -1 & 5 \end{bmatrix}\).

To find: Inverse of matrix \(\begin{bmatrix} 4 & 2 \\\\ -1 & 5 \end{bmatrix}\)

Using the matrix inverse formula,

\( A^{-1} = \dfrac{\text{adj(A)}}{\text{|A|}}\)

\( A^{-1} = \dfrac{1}{det \begin{pmatrix}4 & 2 \\\\ -1 & 5 \end{pmatrix}} \begin{pmatrix}5 & -2 \\\\ 1 & 4 \end{pmatrix}\)

Since, det \(\begin{pmatrix}4 & 2 \\\\ -1 & 5 \end{pmatrix}\) = 22

\( A^{-1} = \dfrac{1}{22} \begin{pmatrix}5 & -2 \\\\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 5/22 & -2/22 \\\\ 1/22 & 4/22 \end{pmatrix} \)

Answer: Inverse of the given matrix \( = \begin{bmatrix} 5/22 & -1/11 \\\\ 1/22 & 2/11 \end{bmatrix}\)

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## Practice Questions on Inverse of Matrix

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## FAQs on Inverse of Matrix

What is the meaning of inverse of matrix.

The inverse of a matrix is another matrix , which multiplies with the given matrix and gives the multiplicative identity. For a matrix A, its inverse is A -1 , and A · A -1 = I. The general formula for the inverse of matrix is equal to the adjoint of a matrix divided by the determinant of a matrix. i.e., A -1 = 1/|A| · Adj A. The inverse of a matrix exists only if the determinant of the matrix is a non-zero value.

## What is the Formula for An Inverse Matrix?

The inverse matrix formula is used to determine the inverse matrix for any given matrix. The inverse of a square matrix, A is A -1 only when: A × A -1 = A -1 × A = I . The inverse matrix formula can be given as, A -1 = adj(A)/|A|; |A| ≠ 0, where A is a square matrix.

## How to Calculate Inverse of Matrix?

The inverse of a square matrix is calculated in 2 simple steps.

- Step 1: First, the determinant and the adjoint of the given square matrix are calculated.
- Step 2: The inverse of the matrix A is equal to \(\dfrac{1}{|A|}\).Adj A.

## How to Find Inverse of a 2 × 2 Matrix?

The inverse of a 2 × 2 matrix is equal to the adjoint of the matrix divided by the determinant of the matrix. For a matrix A = \(\left(\begin{matrix}a&b\\ \\c&d\end{matrix}\right)\), its adjoint is equal to the interchange of the elements of the first diagonal and the sign change of the elements of the second diagonal. The formula for the inverse of the matrix is as follows.

A -1 = \(\dfrac{1}{ad - bc}\left(\begin{matrix}d&-b\\\\-c&a\end{matrix}\right)\)

## How to Use Inverse of Matrix?

The inverse of matrix is useful in solving equations by using the matrix inversion method. The matrix inversion method using the formula of X = A -1 B, where X is the variable matrix, A is the coefficient matrix, and B is the constant matrix.

## Can Inverse of Matrix be Calculated for an Invertible Matrix?

Yes, the inverse of matrix can be calculated for an invertible matrix . The matrix whose determinant is not equal to zero is a non-singular matrix. And for a non-singular matrix, we can find the determinant and the inverse of matrix.

## When Does the Matrix Inverse not Exist in Some Cases?

The inverse of matrix exists only if its determinant value is a non-zero value and when the given matrix is a square matrix. Because the adjoint of the matrix is divided by the determinant of the matrix, to obtain the inverse of a matrix. The matrix whose determinant is a non-zero value is called a non-singular matrix. Matrix inverse is not defined for rectangular matrices .

## Given a 2 × 2 Matrix, What is the Formula for Finding the Inverse of the Matrix?

For a given 2×2 matrix A = \(\left(\begin{matrix}a&b\\ \\c&d\end{matrix}\right)\) , inverse is given by A -1 = \(\dfrac{1}{ad - bc}\left(\begin{matrix}d&-b\\\\-c&a\end{matrix}\right)\). Here A -1 is the inverse of A.

## What is the Inverse of Identity Matrix?

The inverse of an identity matrix is itself. This is because for any identity matrix of order I, we have I × I = I × I = I. For more information, click here .

## How to Use Inverse Matrix Formula?

The inverse matrix formula can be used following the given steps:

- Step 1: Find the matrix of minors for the given matrix.
- Step 2: Then find the matrix of cofactors.
- Step 3: Find the adjoint by taking the transpose of the matrix of cofactors.
- Step 4: Divide it by the determinant.

## What is 3 × 3 Inverse Matrix Formula?

The matrix inverse formula for a 3 × 3 matrix is, A -1 = adj(A)/|A|; |A| ≠ 0 where A = square matrix, adj(A) = adjoint of square matrix, A -1 = inverse matrix.

## What is the Inverse of Diagonal Matrix?

The inverse of a diagonal matrix is again a diagonal matrix in which the elements of the principal diagonal of the matrix inverse are the reciprocals of the corresponding elements of the original matrix. To know how to prove this, click here .

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## How to Solve a System of Equations using Inverse of Matrices?

In mathematics, a matrix is an array of numbers arranged in a rectangular pattern and separated into rows and columns. They’re commonly depicted by enclosing all of the integers within square brackets.

## Determinant

A matrix’s determinant is the scalar value produced for a given square matrix. The determinant is dealt with in linear algebra, and it is computed using the elements of a square matrix. A determinant is a scalar value or number calculated using a square matrix. The square matrix might be 2 × 2, 3 × 3, 4 × 4, or any other form where the number of columns and rows are equal, such as n × n. If S is the set of square matrices, R is the set of integers (real or complex), and f: S → R is defined by f (A) = k, where A ∈ S and k ∈ R, then f (A) is referred to as A’s determinant. A determinant is represented by two vertical lines, i.e., |A|.

Determinant of 2×2 matrix –

## Minors and Cofactors

The matrix created after eliminating the row and column of the matrix in which that specific element lies is defined as the minor of the matrix.

The cofactor of an element in matrix A is produced by multiplying the element’s minor M ij by (-1) i+j . C ij is the symbol for an element’s cofactor. If the minor of a matrix is M ij , then the cofactor of the element would be: C ij = (-1) i+j M ij . The cofactor matrix is the matrix created by the cofactors of the matrix’s components.

## Adjoint of a matrix

Let A=[aij] be an n-dimensional square matrix. A matrix A’s adjoint is the transpose of A’s cofactor matrix. It is symbolized by the letter adj A. Adjoint matrices are sometimes known as adjugate matrices. The adjoint of a square matrix A = [aij]n x n is defined as the transpose of the matrix [Aij]n x n, where Aij is the cofactor of the element aij.

## The inverse of a matrix

A square matrix A is invertible if and only if A is a nonsingular matrix. The inverse of a matrix may be obtained by dividing the adjoint of a matrix by the determinant of the matrix. The inverse of a matrix may be computed by following the steps below:

- Step 1: Determine the minor of the provided matrix.
- Step 2: Convert the acquired matrix into the cofactors matrix.
- Step 3: Finally, the adjugate, and
- Step 4: Multiply it by the determinant’s reciprocal.

Application of Matrices and Determinants

Now, let’s look at how determinants and matrices may be used to solve systems of linear equations in two or three variables and to assess the system’s consistency.

- Consistent System : A system of equations is considered to be consistent if it has (one or more) solutions.
- Inconsistent System : If the solution to a system of equations does not exist, the system is said to be inconsistent.

## Representing linear systems with matrix equations

An augmented matrix can be used to represent a system of equations. Each row in an augmented matrix represents one of the system’s equations, while each column represents a variable or the constant terms. We can see that augmented matrices are a shortcut for formulating systems of equations in this way.

Example: Write the following system of equations as an augmented matrix.

x – 2y = 5

4x – 3y – z = 3

5y – 7z = 9

Let’s write the following matrix in augmented form. If a variable term is not given in the matrix, it is considered that the coefficient of that term is ‘ 0 ‘.

(1)x + (-2)y + (0)z = 5

(4)x + (-3)y + (-1)z = 3

(0)x + (5)y + (-7)z = 9

Solving linear systems with matrix equations

Solving linear equations using a matrix is done by the Matrix method. In this article, we will look at solving linear equations with matrix examples.

## Solving equations with inverse matrices

Matrix Method is used to find the solution of the system of the equations. In the equations, all of the variables should be written in the proper order. On the appropriate sides, write the variables, their coefficients, and constants.

The method of determining the inverse is used to solve a system of linear equations, and it requires two additional matrices. The variables are represented by Matrix X. The constants are represented by Matrix B. Using matrix multiplication, a system of equations with the same number of equations as a variable is defined as,

Let A be the coefficient matrix, X be the variable matrix, and B be the constant matrix to solve a system of linear equations with an inverse matrix. As a result, we’d want to solve the system AX = B. Take a look at the equations below as an example.

Case 1: If A is a nonsingular matrix, it has an inverse.

Let A be the coefficient matrix, X be the variable matrix, and B be the constant matrix to solve a system of linear equations with an inverse matrix. As a result, we’d want to solve the system AX=B. To get the answer, multiply both sides by the inverse of A.

As the inverse of a matrix is unique, this matrix equation offers a unique solution to the given system of equations. The Matrix Method is a method for solving systems of equations.

Case 2: If A is a singular matrix, then | A| = 0. In this case, calculate (adj A) B.

If (adj A) B ≠ O, (O being zero matrices), then the solution does not exist and the system of equations is called inconsistent.

If (adj A) B = O, then the system may be either consistent or inconsistent accordingly as the system has either infinitely many solutions or no solution.

## Sample Problems

- determinant of matrix A
- cofactor matrix A
- adjoint of matrix A
- inverse of matrix A

The given matrix is Determinant of the A = = 3(0+8)+5(-2+4)+3(-4) = 3 × 8 + 5 × 2 + 3 × (-4) = 24 + 10 – 12 units Cofactor of matrix A = C 11 = 0 × (-1) -4 × (-2) = 0 + 8 = 8 C 12 = -((-5) × (-1) -3 × (-2)) = -(5 + 6) = -11 C 13 = (-5) × 4 -3 × 0 = -20 C 21 = −(2 × (-1) -4 × (-1)) = -(-2 + 4) = -2 C 22 = 3 × (-1) -3 × (-1) = -3 + 3 = 0 C 23 = -(3 × 4 – 3 × 2) = -(12 – 6) = -6 C 31 = 2 × (-2) – 0 × (-1) = -4 C 32 = -(3 × (-2) – (-5) × (-1)) = -(-6 – 5) = 11 C 33 = 3 × 0 – (-5) × 2 = 10 Cofactor matrix of A = AdjoinT of matrix A = transpose of cofactor matrix C = Inverse of matrix A = = =

Question 2: Ram is hired for a job with a monthly payment of a specific amount and an annual increase of a predetermined amount. Find his beginning pay and yearly increase if his salary was $300 per month at the end of the first month after 1 year of service and $600 per month at the end of the first month after 3 years of service.

Let “x” and “y” represent the monthly salary and a yearly increase of a certain amount, respectively. According to the question; x + y = 300 ⇢ (i) x + 3y = 600 ⇢ (ii) This can be written as AX = B, where Determinant of A = 1 × 3 – 1 × 1 = 3 – 1 = 2 Adjoin of A = Thus, Using Matrix Inverse, X = A -1 B Therefore; x = $150, y = $150 So, the monthly salary is $150 and the annual increment is $150.

Question 3: The sum of three numbers is 3. If we multiple the second number by 2 and add the first number to it, we get 6. If we multiply the third number by 4 and add the second number to it, we get 10. Represent it algebraically and find the numbers using the matrix method.

Let x, y, and z represent the first, second, and third numbers, respectively. Then, according to the question, we have x + y + z = 3 x + 2y = 6 y + 4z = 10 This can be written as AX = B, where Here, |A|= 1(8 – 0) – 1(4 – 0) + 1(1 – 0) = 8 – 4 + 1 = 5 ≠ 0. Now, find adj A. A 11 = 8 – 0 = 8, A 12 = -(4 – 0) = -4, A 13 = 1 – 0 = 1 A 21 = -(4 – 1) = -3, A 22 = 4 – 0 = 4, A 23 = -(1 – 0) = -1 A 31 = 0 – 2 = -2, A 32 = -(0 – 1) = 1, A 33 = 2 – 1 = 1 Adj. A = Thus, X = A -1 B Therefore;

Question 4: Assume Joe, Max, and Polly went shopping at the mall. Joe pays 45/- for 4 kg of apples, 7 kg of bananas, and 6 kg of guavas, Max pays 30/- for 2 kg of apples and 5 kg of guavas, and Polly pays 35/- for 3 kg of apples, 1 kg of bananas, and 4 kg of guavas. How much do apples, bananas, and guavas cost per kilogram?

Let x, y, and z represent the number of apples, bananas, and guavas, respectively. In accordance to the question: 4x + 7y + 6z = 45 2 x + 5 z = 30 3x + y + 4z = 35 Matrix A contains the kg of apples, bananas, and guavas bought by Joe, Max, and Polly. Matrix B contains the prices paid by the three and matrix X contains the variables. The solution of the given system of equations be X = A -1 B. In order to find the inverse of A, we will first find the determinant of A. Determinant of A = |A| = 4(0 × 4 – 1 × 5) – 7(2 × 4 – 5 × 3) + 6(2 × 1 – 3 × 0) = 4(0 – 5) – 7(8 – 15) + 6(2 – 0) = -20 – 7(-7) + 12 = -20 + 49 + 12 = 41 Adj. of A = The cost of apples per kg = 8.3/- The cost of bananas per kg = 1.1/- The cost of guavas per kg = 2.7/-

Question 5: The cost of 2 kg potatoes, 3 kg tomatoes, and 2 kg flour is 50. The cost of 5 kg potatoes, 1 kg tomatoes and 6 kg flour is 40. The cost of 4 kg potatoes, 6 kg tomatoes and 3 kg flour is 60. Find the cost of each item per kg by the inverse of a matrix.

Let x, y, and z represent the kg of potatoes, tomatoes, and flour, respectively. In accordance to the question: 2x + 3y + 2z = 50 5x + 1y + 6z = 40 4x + 6y + 3z = 60 Matrix A contains the kg of potatoes, tomatoes and flour. Matrix B contains the prices paid and matrix X contains the variables. This can be written as AX = B, where The solution of the given system of equations is X = A -1 B. In order to find the inverse of A, we will first find the determinant of A. Determinant of A |A| = 2(3 – 36) – 3(15 – 24) + 2(30 – 4) = 2 × (-33) – 3(-9) + 2(26) = -66 + 27 + 52 = 13 Now, find the adjoint of A to get the inverse of A. A 11 = 3 – 36 = -33, A 12 = -(15 – 24) = 9, A 13 = 30 – 4 = 26 A 21 = -(9 – 12) = 3, A 22 = 6 – 8 = -2, A 23 = -(12 – 12) = 0 A {-1} = A 31 = 18 – 2 = 16, A 32 = -(12 – 10) = -2, A 33 = 2 – 15 = -13 Thus, X = A -1 B x = 43.8, y = 19.2, z = 40

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## 9-04 Inverse Matrices

- Decompose rational expressions into partial fractions with:
- Find the inverse of a 2×2 matrix.
- Find the inverse of any square matrix.
- Solve a system of equations using an inverse matrix.

SDA NAD Content Standards (2018): PC.6.4

Before talking about inverse matrices, the identity matrix needs to be introduced. The identity matrix is the matrix equivalent of 1 and it given the name I . The identity matrix can be any size square as needed. The downward diagonal elements are all 1 and all other elements are 0. Here are some examples.

$$ \left[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right] \text{or} \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right] \text{or} \left[\begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{matrix}\right] $$

A square matrix multiplied by the identity matrix of the same size will result in the original matrix.

$$ A·I = A $$

An Inverse matrices multiplied with its square matrix will produce the identity matrix. Because the identity matrix is like 1, the inverse matrix is the multiplicative inverse of a square matrix. So, inverse matrices can be used to solve matrix equations by canceling out a matrix.

$$ A·A^{-1} = I $$

## Inverse of a 2×2 Matrix

The inverse of a 2×2 matrix can be found with a formula.

If \(A = \left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), then

$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$

## Example 1: Find the Inverse of a 2×2 Matrix

Find the inverse of \(A = \left[\begin{matrix} 3 & -1 \\ 0 & 2 \end{matrix}\right]\).

Comparing the matrix to \(\left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), a = 3, b = −1, c = 0, and d = 2. Plug those into the formula.

$$ A^{-1} = \frac{1}{3(2)-(-1)(0)} \left[\begin{matrix} 2 & -(-1) \\ -0 & 3 \end{matrix}\right] $$

$$ = \frac{1}{6} \left[\begin{matrix} 2 & 1 \\ 0 & 3 \end{matrix}\right] $$

$$ = \left[\begin{matrix} \frac{1}{3} & \frac{1}{6} \\ 0 & \frac{1}{2} \end{matrix}\right] $$

Find the inverse of \(B = \left[\begin{matrix} 2 & -2 \\ 1 & -3 \end{matrix}\right]\).

## Find the Inverse of a Square Matrix

To find the inverse of a square matrix, A

- Create an augmented matrix with the identity matrix \([A : I]\)
- Use Gauss-Jordan Elimination to change the left side into the identity matrix. The right side will then be the inverse, \([I : A^{-1}]\)

## Example 2: Find an Inverse Matrix

Find the inverse of \(A = \left[\begin{matrix} 2 & 0 & 1 \\ 0 & -1 & -3 \\ -2 & 4 & 3 \end{matrix}\right]\).

Create an augmented matrix with the identity matrix.

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ -2 & 4 & 3 & : & 0 & 0 & 1 \end{matrix}\right] $$

Use Gauss-Jordan Elimination to turn the left side into the identity matrix. There is already a 0 in the 1st column on the 2nd row, so go straight to getting rid of the −2 in the 3rd row by adding the 1st row to the 3rd row.

When showing your work on your assignment, you typically only show the steps with the red numbers in them.

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ \color{blue}{-2} & 4 & 3 & : & 0 & 0 & 1 \end{matrix}\right] $$

$$ \begin{matrix} \swarrow \\ \downarrow \\ +\rightarrow \end{matrix} \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ \color{red}{0} & \color{red}{4} & \color{red}{4} & : & \color{red}{1} & \color{red}{0} & \color{red}{1} \end{matrix}\right] $$

Now work down the 2nd column. Get rid of the 4 in the 3rd row by multiplying the 2nd row by 4 and add to the 3rd row.

$$ \begin{matrix} \quad \\ \swarrow ×4 \\ +\rightarrow \end{matrix} \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ 0 & \color{blue}{4} & 4 & : & 1 & 0 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ \color{red}{0} & \color{red}{0} & \color{red}{-8} & : & \color{red}{1} & \color{red}{4} & \color{red}{1} \end{matrix}\right] $$

The matrix is almost in row echelon form except for the leading nonzero entry in the 2nd and 3rd rows is not 1. If they were turned into a 1, then there would be fractions. However, fractions would make the remaining process a bit more of a nuisance, so they will be left for now. It is time to begin up, start getting zeros in the 3rd column and work from bottom up. Get rid of the −3 in the 2nd row by multiplying the 3rd row by −3 and adding to 8 times the 2nd row.

$$ \begin{matrix} \quad \\ +\rightarrow ×8 \\ \nwarrow ×\left(-3\right) \end{matrix} \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & \color{blue}{-3} & : & 0 & 1 & 0 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{-8} & \color{red}{0} & : & \color{red}{-3} & \color{red}{-4} & \color{red}{-3} \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

Continue working up the the 3rd column. Get rid of the 1 in the 1st row adding the 3rd row to 8 times the 1st row.

$$ \begin{matrix} +\rightarrow ×8 \\ \uparrow\qquad \\ \nwarrow \quad \end{matrix} \left[\begin{matrix} 2 & 0 & \color{blue}{1} & : & 1 & 0 & 0 \\ 0 & -8 & 0 & : & -3 & -4 & -3 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{16} & \color{red}{0} & \color{red}{0} & : & \color{red}{9} & \color{red}{4} & \color{red}{1} \\ 0 & -8 & 0 & : & -3 & -4 & -3 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

Now that the 3rd column is done, move up the 2nd column. However, there is already a 0 in the 1st row of the 2nd column, so now just multiply to get leading 1's.

$$ \begin{matrix} ×\frac{1}{16} \\ ×\left(-\frac{1}{8}\right) \\ ×\left(-\frac{1}{8}\right) \end{matrix} \left[\begin{matrix} 16 & 0 & 0 & : & 9 & 4 & 1 \\ 0 & -8 & 0 & : & -3 & -4 & -3 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{0} & : & \color{red}{\frac{9}{16}} & \color{red}{\frac{1}{4}} & \color{red}{\frac{1}{16}} \\ \color{purple}{0} & \color{purple}{1} & \color{purple}{0} & : & \color{purple}{\frac{3}{8}} & \color{purple}{\frac{1}{2}} & \color{purple}{\frac{3}{8}} \\ \color{green}{0} & \color{green}{0} & \color{green}{1} & : & \color{green}{-\frac{1}{8}} & \color{green}{-\frac{1}{2}} & \color{green}{-\frac{1}{8}} \end{matrix}\right] $$

The inverse is the right side.

$$ A^{-1} = \left[\begin{matrix} \frac{9}{16} & \frac{1}{4} & \frac{1}{16} \\ \frac{3}{8} & \frac{1}{2} & \frac{3}{8} \\ -\frac{1}{8} & -\frac{1}{2} & -\frac{1}{8} \end{matrix}\right] $$

Find the inverse of \(B = \left[\begin{matrix} 3 & 1 & 0 \\ 1 & 2 & -2 \\ 0 & 1 & 4 \end{matrix}\right]\).

\(B^{-1} = \left[\begin{matrix} \frac{5}{13} & -\frac{2}{13} & -\frac{1}{13} \\ -\frac{2}{13} & \frac{6}{13} & \frac{3}{13} \\ \frac{1}{26} & -\frac{3}{26} & \frac{5}{26} \end{matrix}\right]\).

## Use an Inverse Matrix to Solve a System of Equations

Inverse matrices can be used to solve a system of equations. This is done to by writing the system of equations as a matrix equation. Matrix A is made of the coefficients, B is the constants, and X is the variables. This is complicated because dividing by a matrix is not defined, so instead multiply by the inverse.

$$ A·X = B $$

You cannot divide by a matrix, so multiply both sides by the inverse. Remember order is important when performing matrix multiplication. To get the inverse to multiply with matrix A , the inverse had to go to the far left, so the inverse also has to go to the left of the right side of the equation.

$$ \color{blue}{A^{-1}} ·A·X = \color{blue}{A^{-1}}·B $$

Simplify by remembering \(A^{-1}·A = I\).

$$ \color{blue}{I}·X = A^{-1}·B $$

Simplify by using \(I·X = X\).

$$ \color{blue}{X} = A^{-1}·B $$

In summary, if you start with \(A·X = B\), then the solution is \(X = A^{-1}·B\). For a system of equations, A was the coefficients and B was the constants. The solution is simply the inverse of the coefficient matrix multiplied with the constants matrix.

- Write the system as a matrix equation \(A·X = B\) where A is the coefficients, X is the variables, and B is the constants.
- Find the inverse of the coefficient matrix, \(A^{-1}\).
- The solution is \(X = A^{-1}·B\), just multiply the inverse with the constant matrix.

## Example 3: Use an Inverse Matrix to Solve a System of Equations

Solve using an inverse matrix \(\left\{\begin{align} 2x - 6y &= 5 \\ 8x + 4y &= -1 \end{align}\right.\)

Write the system as a matrix equation. [coefficients] × [variables] = [constants]

$$ \left[\begin{matrix} 2 & -6 \\ 8 & 4 \end{matrix}\right] \left[\begin{matrix} x \\ y \end{matrix}\right] = \left[\begin{matrix} 5 \\ -1 \end{matrix}\right] $$

Find the inverse of the coefficient matrix, \(\left[\begin{matrix} 2 & -6 \\ 8 & 4 \end{matrix}\right]\). Comparing the matrix to \(\left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), a = 2, b = -6, c = 8, and d = 4. Plug those into the formula.

$$ A^{-1} = \frac{1}{2(4)-(-6)(8)} \left[\begin{matrix} 4 & 6 \\ -8 & 2 \end{matrix}\right] $$

$$ A^{-1} = \frac{1}{56} \left[\begin{matrix} 4 & 6 \\ -8 & 2 \end{matrix}\right] $$

Multiply the inverse matrix with the constant matrix, \(X = A^{-1}·B\). See lesson 9-03 for multiplying matrices.

$$ \left[\begin{matrix} x \\ y \end{matrix}\right] = \frac{1}{56} \left[\begin{matrix} 4 & 6 \\ -8 & 2 \end{matrix}\right] \left[\begin{matrix} 5 \\ -1 \end{matrix}\right] $$

$$ \left[\begin{matrix} x \\ y \end{matrix}\right] = \frac{1}{56} \left[\begin{matrix} 14 \\ -42 \end{matrix}\right] $$

$$ \left[\begin{matrix} x \\ y \end{matrix}\right] = \left[\begin{matrix} \frac{1}{4} \\ -\frac{3}{4} \end{matrix}\right] $$

The solution is \(\left(\frac{1}{4}, -\frac{3}{4}\right)\).

Solve using an inverse matrix \(\left\{\begin{align} 2x + 7y &= 67 \\ 3x - 4y &= -56 \end{align}\right.\)

## Lesson Summary

Helpful videos about this lesson.

- Mr. Wright Teaches the Lesson
- The Identity Matrix
- Determining Inverse Matrices
- Using a Matrix Equation to Solve a System of Equations

## Practice Exercises

- Multiply the matrices and determine if they are inverses. (Hint: The product should be the identity matrix.) \(\left[\begin{matrix} 2 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 2 & 0 \end{matrix}\right]\) and \(\left[\begin{matrix} 0.5 & 0.5 & 0 \\ 0 & 0 & 0.5 \\ 0 & -1 & 0 \end{matrix}\right]\)

Find the inverse matrix.

- \(\left[\begin{matrix} 2 & -1 \\ 3 & 1 \end{matrix}\right]\)
- \(\left[\begin{matrix} 1 & 4 \\ -2 & 5 \end{matrix}\right]\)
- \(\left[\begin{matrix} 2 & -1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & -1 \end{matrix}\right]\)
- \(\left[\begin{matrix} 1 & 3 & -2 \\ 1 & -2 & 1 \\ 4 & 0 & 2 \end{matrix}\right]\)
- \(\left[\begin{matrix} 3 & 2 & -3 \\ 0 & -4 & 2 \\ -3 & 0 & 1 \end{matrix}\right]\)

Use inverse matrices to solve the system of equations.

- \(\left\{\begin{align} 2x + 2y &= -2 \\ x - 5y &= 8 \end{align}\right.\)
- \(\left\{\begin{align} x - 3y &= -6 \\ 2x + y &= -5 \end{align}\right.\)
- \(\left\{\begin{align} 3x + 2y - 3z &= 4 \\ -4y + 2z &= -6 \\ -3x + z &= -2 \end{align}\right.\)

Problem Solving

- Fred is investing $5000 in two different accounts to stay diversified. One account pays 6% simple interest and the other pays 9% simple interest. Use an inverse matrix to determine how much he should invest in each account to earn an average of 8%.

Mixed Review

- (9-03) \(\left[\begin{matrix} 2 & 5 & -7 \end{matrix}\right] - 3 \left[\begin{matrix} -1 & 10 & 3 \end{matrix}\right]\)
- (9-03) \(\left[\begin{matrix} 2 & 1 & -2 \end{matrix}\right] \left[\begin{matrix} 1 & 0 \\ -1 & -2 \\ 0 & 5 \end{matrix}\right]\)
- (9-01) Add 2 times row 1 to row 3 and identify what it accomplished. \(\left[\begin{matrix} 1 & 3 & 0 & 2 \\ 0 & 3 & -2 & -5 \\ -2 & 8 & -6 & 2 \end{matrix}\right]\)
- (6-03) If \(\overset{\rightharpoonup}{m} = \langle 2, -1 \rangle\) and \(\overset{\rightharpoonup}{n} = \langle -3, 1 \rangle\), find \(2 \overset{\rightharpoonup}{m} - \overset{\rightharpoonup}{n}\).
- (5-04) Solve \(2 \sin x − \sqrt{2} = 0\) on the interval \(0 ≤ x < 2π\).
- Show work, yes they are inverses
- \(\left[\begin{matrix} \frac{1}{5} & \frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{matrix}\right]\)
- \(\left[\begin{matrix} \frac{5}{13} & -\frac{4}{13} \\ \frac{2}{13} & \frac{1}{13} \end{matrix}\right]\)
- \(\left[\begin{matrix} \frac{1}{2} & \frac{1}{6} & \frac{1}{6} \\ 0 & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & -1 \end{matrix}\right]\)
- \(\left[\begin{matrix} \frac{2}{7} & \frac{3}{7} & \frac{1}{14} \\ -\frac{1}{7} & -\frac{5}{7} & \frac{3}{14} \\ -\frac{4}{7} & -\frac{6}{7} & \frac{5}{14} \end{matrix}\right]\)
- \(\left[\begin{matrix} -\frac{1}{3} & -\frac{1}{6} & -\frac{2}{3} \\ -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -1 & -\frac{1}{2} & -1 \end{matrix}\right]\)
- \(\left(\frac{1}{2}, -\frac{3}{2}\right)\)
- $1666.67 at 6%, $3333.33 at 9%
- \(\left[\begin{matrix} 5 & -25 & -16 \end{matrix}\right]\)
- \(\left[\begin{matrix} 1 & -12 \end{matrix}\right]\)
- \(\left[\begin{matrix} 1 & 3 & 0 & 2 \\ 0 & 3 & -2 & -5 \\ 0 & 14 & -6 & 6 \end{matrix}\right]\); Made a 0 in the 3rd row.
- \(\langle 7, -3 \rangle\)

## IMAGES

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