If A is a non-singular square matrix, there is an existence of n x n matrix A^-1, which is called the inverse of a matrix A such that it satisfies the property:

AA^-1 = A^-1A = I, where I is  the Identity matrix

The identity matrix for the 2 x 2 matrix is given by

\(I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\)

It is noted that in order to find the matrix inverse, the square matrix should be non-singular whose determinant value does not equal to zero.

Let us take the square matrix A

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

Where a, b, c, and d represents the number.

The determinant of the matrix A is written as ad-bc, where the value is not equal to zero. In this article, let us discuss the important properties of matrices inverse with example.

Also, read:

• Types Of Matrices
• Determinants and Matrices
• Determine The Order Of Matrix
• Application Of Matrices

Matrix Inverse Properties

The list of properties of matrices inverse is given below. Go through it and simplify the complex problems.

If A and B are the non-singular matrices, then the inverse matrix should have the following properties

• (A^-1)^-1 =A
• (AB)^-1 =A^-1B^-1
• (ABC)^-1 =C^-1B^-1A^-1
• (A₁ A₂….A_n)^-1 =A_n^-1A_n-1^-1……A₂^-1A₁^-1
• (A^T)^-1 =(A^-1)^T
• (kA)^-1 = (1/k)A^-1
• AB = I_n, where A and B are inverse of each other.
• If A is a square matrix where n>0, then (A^-1)ⁿ =A^-n

Where A^-n = (Aⁿ)^-1

Solved Example

The example of finding the inverse of the matrix is given in detail. Go through it and learn the problems using the properties of matrices inverse.

Question:

Find the inverse A^-1 of the matrix \(A=\begin{bmatrix} 2 & 1 & 1\\ 3 & 2 & 1\\ 2 & 1 & 2 \end{bmatrix}\)

Solution:

Given: \(A=\begin{bmatrix} 2 & 1 & 1\\ 3 & 2 & 1\\ 2 & 1 & 2 \end{bmatrix}\)

We know that

A^-1 = adj (A) / det(A)

A^-1 = adj (A) / |A|

\(|A|=2\begin{vmatrix} 2 & 1\\ 1 & 2 \end{vmatrix}-1\begin{vmatrix} 3 & 1\\ 2 & 2 \end{vmatrix}+1\begin{vmatrix} 3 & 2\\ 2 & 1 \end{vmatrix}\)

|A| = 2(4-1) -1(6-2)+1(3-4)

|A| = 2(3) -1(4)+1(-1)

|A| = 6-4-1

|A| = 1

Now, find Adj(A):

\(A_{c}=\begin{bmatrix} (4-1) & -(6-2) & (3-4) \\ -(2-1) &(4-2) &-(2-2) \\ (1-2) &-(2-3) & (4-3) \end{bmatrix}\) \(A_{c}=\begin{bmatrix} 3 & -4 & -1 \\ -1 &2 &0 \\ -1 &1 & 1 \end{bmatrix}\)

Now, take the transpose of the cofactor matrix

\(A_{c}^{T}=\begin{bmatrix} 3 & -1 & -1 \\ -4 &2 &1 \\ -1 &0 & 1 \end{bmatrix}\)

Therefore,

A^-1 = adj (A) / |A|

\(A^{-1}=\frac{1}{1}\begin{bmatrix} 3 & -1 & -1 \\ -4 &2 &1 \\ -1 &0 & 1 \end{bmatrix}\) \(A^{-1}=\begin{bmatrix} 3 & -1 & -1 \\ -4 &2 &1 \\ -1 &0 & 1 \end{bmatrix}\)

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