How To Find Inverse Of A Matrix

Table of Contents

In the previous article, you learned about inverse of matrix and why it is important. You will learn how to find inverse of a matrix in this article. There are two primary method of finding inverse of any square invertible matrix – Classical adjoint method and Gauss -Jordan elimination method.

Methods To Find Inverse Of Matrix

Classical Adjoint Method

If is an invertible matrix then we can find the inverse of matrix with the adjoint of matrix .

But,before we begin you must understand a few terminologies.

Determinant – It is a special number obtained from a square matrix, non-square matrix do not have determinants. If is a square matrix then there is a number of ways to denote its determinant.

\begin{aligned} &A = \begin{bmatrix}a & b\\c & d\end{bmatrix}\\\\ &det(A) \hspace{5px} or \hspace{5px}\begin{vmatrix}A\end{vmatrix} \hspace{5px} or \hspace{5px}\begin{vmatrix}a & b\\c & d\end{vmatrix}\\\\ &det(A)= ad - bc \end{aligned}

Minor of a matrix – The minor of a square matrix is determinant obtained by deleting a row and a column from the determinant of a larger square matrix. It is denoted by for element where is the i^throw and is the j^th column.

The determinant of 2 x 2 matrix.

\begin{aligned} &det(A) =\begin{vmatrix}a & b\\c & d\end{vmatrix}\\\\ &M_{11} = \begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}d\end{vmatrix}= d\\\\ &M_{12} = \begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}c\end{vmatrix}= c\\\\ &M_{21} = \begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}b\end{vmatrix}= b\\\\ &M_{22} = \begin{vmatrix}a & b\\c & d\end{vmatrix} = \begin{vmatrix}a\end{vmatrix}= a\\\\ \end{aligned}

You get following matrix of determinant.

= \begin{bmatrix}d & c\\b & a\end{bmatrix}

Cofactor of a Matrix – The cofactor of a matrix can be found from determinant of minors by assigning appropriate negative or positive signs. Each cofactor can be found by following equation.

\begin{aligned} &A_{ij}= (-1)^{i+j}M_{ij}\\\\ &A_{11}= (-1)^2 \times det(d) = d\\\\ &A_{12}= (-1)^3 \times det(c) = -c\\\\ &A_{21}= (-1)^3 \times det(b) = -b\\\\ &A_{22}= (-1)^4 \times det(a) = a \end{aligned}

Adjoint of a Matrix – The adjoint of a matrix is obtained by transposing the co-factor matrix. It is denoted as .

\begin{aligned} adj A = \left(A_{ij}\right)^T = \begin{bmatrix}d & -b\\-c & a\end{bmatrix} \end{aligned}

Finding Inverse Of Matrix Through Adjoint Method

The process of finding inverse of matrix using adjoint method is as follows.

Find the matrix of minors
Find the matrix of co-factors
Find the determinant det(A) by multiplying first row of matrix A with first row of co-factor matrix of A.
Find the adjoint matrix usign co-factor matrix of A.
Multiply 1/det(A) with adjoint of A (adj A) to get the inverse matrix of A.

We will find inverse of a matrix using the adjoint of matrix in the next section. First we must find the inverse of matrix, then and finally matrix.

Example #1 : Find the inverse of matrix using adjoint method.

A = \begin{bmatrix}3\end{bmatrix}

Solution:

Step 1: There is no minor for 1×1 matrix.

Step 2: There is no cofactor for 1×1 matrix.

Step 3: The determinant of A is .(invertible)

Step 4: The adjoint of matrix A is .

To compute the inverse of matrix A use

\begin{aligned} &A^{-1} = 1/det(A) \times \begin{bmatrix}1\end{bmatrix}\\\\ &A^{-1} = 1/3 \times \begin{bmatrix}1\end{bmatrix}\\\\ &A^{-1} = 1/3 \end{aligned}

Example #2 : Find the inverse of matrix using adjoint method.

B = \begin{bmatrix}2 & 1\\3 & 7\end{bmatrix}

Solution:

Let be a square and invertible matrix of order .

Step 1: The minors of matrix are

\begin{aligned} &M_{11} = \begin{vmatrix}7\end{vmatrix}= 7\\\\ &M_{12} = \begin{vmatrix}3\end{vmatrix}= 3\\\\ &M_{21} = \begin{vmatrix}1\end{vmatrix}= 1\\\\ &M_{22} = \begin{vmatrix}2\end{vmatrix}= 2 \end{aligned}

The minor matrix is,

Step 2: The cofactor of matrix is obtained from minor matrix.

\begin{aligned} &B_{11} = (-1)^{1+1}M_{11} = 7\\\\ &B_{12} = (-1)^{1+2}M_{12} = -3\\\\ &B_{21} = (-1)^{2+1}M_{21} = -1\\\\ &B_{22} = (-1)^{2+2}M_{22} = 2 \end{aligned}

The cofactor matrix is,

= \begin{bmatrix}7 & -3\\-1 & 2\end{bmatrix}

Step 3: To compute the determinant simply multiply corresponding elements of top row of matrix and tow row of cofactor of and add them.

Top row of matrix B = 2, 1

Top row of cofactor matrix = 7 -3

det(B) = 2 \times 7 + 1 \times (-3) = 14 - 3 = 11

Step 4: Find the adjoint of the matrix by transposing cofactor matrix.

\begin{aligned} &cofactor \hspace{1ex} of \hspace{1ex} B = \begin{bmatrix}7 & -3\\-1 & 2\end{bmatrix}\\\\ &adj B = \begin{bmatrix}B_{ij}\end{bmatrix}^T\\\\ &adj B = \begin{bmatrix}7 & -1\\-3 & 2\end{bmatrix} \end{aligned}

To find the inverse of matrix use following.

\begin{aligned} &B^{-1} = 1/11 \times \begin{bmatrix}7 & -1\\-3 & 2\end{bmatrix}\\\\ &B^{-1} = \begin{bmatrix}7/11 & -1/11\\-3/11& 2/11\end{bmatrix} \end{aligned}

Verify the inverse of matrix .

\begin{aligned} &BB^{-1} = B^{-1}B = I\\\\ &= \begin{bmatrix}2 & 1\\3 & 7\end{bmatrix} \times \begin{bmatrix}7/11 & -1/11\\-3/11& 2/11\end{bmatrix}\\\\ &= \begin{bmatrix}14/11 + -3/11& -2/11+ 2/11\\21/11 + -21/11& -3/11+ 14/11\end{bmatrix}\\\\ &= \begin{bmatrix}11/11&0\\0&11/11\end{bmatrix}\\\\ &= \begin{bmatrix}1&0\\0&1\end{bmatrix} \end{aligned}

Example #3 : Find the inverse of 3 x 3 matrix using the adjoint method.

A = \begin{bmatrix}1 & -2 & 1\\6 & 2 & 3\\2 & 1 & 4\end{bmatrix}

Solution :

Given the matrix of order .

A = \begin{bmatrix}1 & -2 & 1\\6 & 2 & 3\\2 & 1 & 4\end{bmatrix}

Step 1: Find the minors of the matrix .

\begin{aligned} &M_{11} = \begin{vmatrix}a_{22}.a_{33}-a_{23}.a_{32}\end{vmatrix} = \begin{vmatrix}8 - 3\end{vmatrix} = 5\\\\ &M_{12} = \begin{vmatrix}a_{21}.a_{33}-a_{23}.a_{31}\end{vmatrix} = \begin{vmatrix}24 - 6 \end{vmatrix}= 18\\\\ &M_{13} = \begin{vmatrix}a_{21}.a_{32}-a_{22}.a_{31}\end{vmatrix} = \begin{vmatrix}6 - 4\end{vmatrix} = 2\\\\ &M_{21} = \begin{vmatrix}a_{12}.a_{33}-a_{13}.a_{32}\end{vmatrix} = \begin{vmatrix}-8 - 1\end{vmatrix} = -9\\\\ &M_{22} = \begin{vmatrix}a_{11}.a_{33}-a_{13}.a_{31}\end{vmatrix} = \begin{vmatrix}4 - 2\end{vmatrix} = 2\\\\ &M_{23} = \begin{vmatrix}a_{11}.a_{32}-a_{12}.a_{31}\end{vmatrix} = \begin{vmatrix}1 + 4 \end{vmatrix}= 5\\\\ &M_{31} = \begin{vmatrix}a_{12}.a_{23}-a_{13}.a_{22}\end{vmatrix} = \begin{vmatrix}-6 - 2\end{vmatrix} = -8\\\\ &M_{32} = \begin{vmatrix}a_{11}.a_{23}-a_{13}.a_{21}\end{vmatrix} = \begin{vmatrix}3 - 6\end{vmatrix} = -3\\\\ &M_{33} = \begin{vmatrix}a_{11}.a_{22}-a_{12}.a_{21}\end{vmatrix} = \begin{vmatrix}2 + 12\end{vmatrix} = 14 \end{aligned}

We have the following matrix of minors.

\begin{bmatrix}5 & 18 & 2\\-9 & 2 & 5\\-8 & -3 & 14\end{bmatrix}

Step 2: Find the cofactors of matrix .

We can use the matrix of minors to find the matrix of cofactors.

\begin{aligned} &A_{11} = (-1)^2 . 5 = 5\\ &A_{12} = (-1)^3 . 18 = -18\\ &A_{13} = (-1)^4 . 2 = 2\\\\ &A_{21} = (-1)^3 . -9 = 9\\ &A_{22} = (-1)^4 . 2 = 2\\ &A_{23} = (-1)^5 . 5 = -5\\\\ &A_{31} = (-1)^4 . -8 = -8\\ &A_{32} = (-1)^5 . -3 = 3\\ &A_{33} = (-1)^4 . 14 = 14 \end{aligned}

We get the cofactor matrix of .

= \begin{bmatrix}5 & -18 & 2\\9 & 2 & -5\\-8 & 3 & 14\end{bmatrix}

Step 3: Find determinant of the matrix .

Top row of matrix A = 1, -2, 1

Top row of coactor of A = 5, -18, 2

det(A) = (1)(5)+ (-2)(-18) + (1)(2) = 5 + 36 + 2 = 43

Step 4: Find the adjoint of matrix . The adjoint of the matrix can be obtained from transposing the cofactor matrix.

\begin{aligned} &Adj A = \begin{bmatrix}A_{ij}\end{bmatrix}^T = \begin{bmatrix} 5 & -18 & 2\\9 & 2 & -5\\-8 & 3 & 14\end{bmatrix}^T\\\\ &Adj A = \begin{bmatrix} 5 & 9 & -8\\-18& 2 & 3\\2 & -5 & 14\end{bmatrix} \end{aligned}

Step 5: Find the inverse of matrix using following equation.

\begin{aligned} &A^{-1} = 1/det(A) \times Adj A\\\\ &A^{-1} = 1/43 \times \begin{bmatrix} 5 & 9 & -8\\-18& 2 & 3\\2 & -5 & 14\end{bmatrix}\\\\ &A^{-1} = \begin{bmatrix}5/43& 9/43& -8/ 43\\-18/43& 2/43& 3/43\\2/43& -5/43&14/43\end{bmatrix} \end{aligned}

Verify that .

Verify results:

\begin{aligned} &= \begin{bmatrix}1 & -2 & 1\\6 & 2 & 3\\2 & 1 & 4\end{bmatrix} \times \begin{bmatrix}5/43&9/ 43& -8/ 43\\-18/43& 2/43& 3/43\\2/43& -5/43&14/43\end{bmatrix}\\\\ & = \begin{bmatrix}5/43+ 36/43+2/43& 9/43 +( -4)/43 + (-5)/43& -8/43+ 6/43+14/43\\30/43+( -36)/43+ 6/43& 54/43+4/43+ (-15)/43& -48/43+6/43+42/43\\10/43+(-18)/43+ 8/43&18/43+2/43+ (-20)/43& -16/43+3/43+56/43\end{bmatrix}\\\\ & =\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix} \end{aligned}

Finding Inverse of Matrix Using Gauss-Jordan Elimination Method

Advertisements

The Gauss-Jordan elimination method convert a matrix into reduced-row echelon form to find the value of solution vector in , but here we use the technique to find inverse matrix .

Row Operations

The Gauss-Jordan technique involves row operations on augmented matrix obtained from the system of linear equations that transforms the matrix into identity matrix where is the order of matrix .

These row operations are

Multiply a row with scalar where .
Interchange row with row .
Addition of times row to row .

Example #5 : Row operations

Let be matrix of order .

A = \begin{bmatrix}1 & 3\\2 & 1\end{bmatrix}

Multiply a row 1 with 3.

R_1 \times 3 = \begin{bmatrix}3 & 9\\2 & 1\end{bmatrix}

Interchange row 2 with row 1.

R_2 \Leftrightarrow R_1 = \begin{bmatrix}2 & 1\\3 & 9\end{bmatrix}

Add 2 times row 1 to row 2.

2R_1 + R_2 = \begin{bmatrix}2 & 1\\7 & 11\end{bmatrix}

Elementary Matrix

An elementary matrix is a matrix obtained from performing a single row operation on identity matrix .

Suppose A is a matrix of order , then the product is same as performing that row operation on matrix .

Example #6 : Elementary Matrix

Suppose is a matrix of order .

A = \begin{bmatrix}1 & 5 & 8\\2 & 1 & 3\\0 & 1 & 6\end{bmatrix}

Let be row operation on of order .

Let \hspace{5px} E = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\2 & 0 & 1\end{bmatrix}

Let us perform on .

= \begin{bmatrix}1 & 5 & 8\\2 & 1 & 3\\2 & 11 & 22\end{bmatrix} \hspace{5px} (1)

Now, let us find matrix.

\begin{aligned} &EA = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\2 & 0 & 1\end{bmatrix} \times \begin{bmatrix}1 & 5 & 8\\2 & 1 & 3\\0 & 1 & 6\end{bmatrix}\\\\ &EA = \begin{bmatrix}1+0+0 & 5+0+0 & 8+0+0\\0+2+0 & 0+1+0 & 0+3+0\\2+0+0 & 10+0+1 & 16+0+6\end{bmatrix}\\\\ &EA = \begin{bmatrix}1 & 5 & 8\\2 & 1 & 3\\2 & 11 & 22\end{bmatrix}\hspace{5px} (2) \end{aligned}

Therefore, (1) and (2) are same.

Inverse Operations

If matrix is a result of row operation on identity matrix , then there exists some operation if performed on will give back . Such an operation is called an Inverse operation.

For every elementary row operation, there is an equivalent inverse operation. Check the table below.

Elementary Row Operation	Inverse Operation
Multiply row i by c where c != 0	Multiply row i by 1/c
Interchange row i with row j	Interchange row j with row i
Add k times row i to row j	Add -k times row i to row j

Inverse operations

Example #7 : Inverse Operations

Multiply row by .

\begin{aligned} &E = k \times R_1 = 3 \times \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\\\\ &E = \begin{bmatrix}3 & 0\\0 & 1\end{bmatrix} \end{aligned}

Multiply row by where

\begin{aligned} &1/3 \times R_1 = 1/3\times \begin{bmatrix}3 & 0\\0 & 1\end{bmatrix}\\\\ &I = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \end{aligned}

From the above we can conclude that

1. Some operation on give the elementary matrix .

2. The inverse of previous operation on E will give back .

3. Inverse operation on will result in an inverse matrix such that

E.E^{-1} = E^{-1}.E = I

Note that is invertible matrix and has determinant greater than 0.

4 Necessary Statements For Gauss-Jordan technique and inverses

Before we find inverse of a matrix using Gauss-Jordan technique, there are 4 necessary prepositions that must be true about the matrix and its inverse.

If matrix is square matrix these 4 statement must always be true.

Matrix is invertible meaning the determinant is greater than 0 and does not have a zero row.
has only trivial solution.
The matrix is reduced to reduced-row echelon form which is identity matrix . This is because has trivial solution only, which means has one solution for every .