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# Inverse of 2x2 Matrix

In mathematics, a matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix. Some matrices may contain inverse and some may not. Thus, the matrix inverse can be defined as “If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A” and it is denoted by A-1. In that case A is said to be invertible. In this article, you will learn how to find the inverse of a 2 x 2 matrix using different methods.

## How to Find Inverse of 2×2 Matrix

To find the inverse of any matrix, it is important to observe that the determinant of the matrix should not be 0. If the matrix determinant is equal to zero, then the inverse of that matrix does not exist.

For an invertible matrix of order 2 x2, we can find the inverse in two different methods such as:

• Inverse using Elementary operations
• Using the Inverse matrix formula

In the next section, you will go through the examples on finding the inverse of given 2×2 matrices.

### Inverse of a 2×2 Matrix Using Elementary Row Operations

If A is a matrix such that A-1 exists, then to find the inverse of A, i.e. A-1 using elementary row operations, write A = IA and apply a sequence of row operations on A = IA till we get I = BA. The matrix B will be the inverse of A. Similarly, if to find A-1 using column operations, then write A = AI and implement a sequence of column operations on A = AI until we get AB = I. Let’s have a look at the below example to understand how we can find the inverse of a given 2×2 matrix using elementary row operations.

Example 1: Find the inverse of the matrix

$$\begin{array}{l}A=\begin{bmatrix} 1 & 2\\ 2& -1 \end{bmatrix}\end{array}$$
using elementary row operations.

Solution:

Given,

$$\begin{array}{l}A=\begin{bmatrix} 1 & 2\\ 2& -1 \end{bmatrix}\end{array}$$

To apply elementary row operations, let’s write the given matrix in the form of A = IA.

Thus,

$$\begin{array}{l}\begin{bmatrix} 1 & 2\\ 2& -1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\end{array}$$

Applying R2 → R2 – 2R1,

$$\begin{array}{l}\begin{bmatrix} 1 & 2\\ 0& -5 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ -2 & 1 \end{bmatrix}A\end{array}$$

Now, applying R2 → (-1/5)R2 we get,

$$\begin{array}{l}\begin{bmatrix} 1 & 2\\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ \frac{2}{5}& \frac{-1}{5} \end{bmatrix}A\end{array}$$

Applying R1 → R1 – 2R2 to get the identity matrix on LHS,

$$\begin{array}{l}\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{5} & \frac{2}{5}\\ \frac{2}{5}& \frac{-1}{5} \end{bmatrix}A\end{array}$$

This is of the form I = A-1A

Therefore,

$$\begin{array}{l}A^{-1}=\begin{bmatrix} \frac{1}{5} & \frac{2}{5}\\ \frac{2}{5}& \frac{-1}{5} \end{bmatrix}\end{array}$$

Alternatively, we can also use elementary column operations to get the inverse for the given matrix.

### Inverse of 2×2 Matrix Formula

Let

$$\begin{array}{l}A=\begin{bmatrix} a &b \\ c &d \end{bmatrix}\end{array}$$
be the 2 x 2 matrix. The inverse of matrix A can be found using the formula given below.

$$\begin{array}{l}A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\ -c & a \end{bmatrix}\end{array}$$

Here, ad – bc = det(A) {determinant of the matrix A}

And

$$\begin{array}{l}\begin{bmatrix} d &-b \\ -c & a \end{bmatrix}\end{array}$$
is the adjoint of matrix A.

Go through the example given below to understand how to find the 2×2 matrix’s inverse using the formula.

Example 2: Find the inverse of

$$\begin{array}{l}A=\begin{bmatrix} 3 & 10\\ 2& 7 \end{bmatrix}\end{array}$$
using the formula.

Solution:

Given,

$$\begin{array}{l}A=\begin{bmatrix} 3 & 10\\ 2& 7 \end{bmatrix}\end{array}$$

Now, we have to find |A|.

$$\begin{array}{l}|A|=\begin{vmatrix} 3 & 10\\ 2& 7 \end{vmatrix}\end{array}$$

det(A) = 3(7) – 2(10) = 21 – 20 = 1 ≠ 0

$$\begin{array}{l}\begin{bmatrix} 7 & -10\\ -2& 3 \end{bmatrix}\end{array}$$
$$\begin{array}{l}A^{-1}=\frac{1}{1}\begin{bmatrix} 7 & -10\\ -2& 3 \end{bmatrix}\\ Therefore, \ A^{-1}=\begin{bmatrix} 7 & -10\\ -2& 3 \end{bmatrix}\\\end{array}$$