# How to Find Inverses of 2×2 Matrices?

In matrices, the inverse of a matrix $$A$$ is a matrix that when multiplied by $$A$$ gives the identity matrix. Here, you learn about how to find inverses of $$2×2$$ matrices. If $$A$$ is a non-singular square matrix, there is an existence of $$n x n$$ matrix $$A^{-1}$$, which is called the inverse matrix of $$A$$ such that it satisfies the property:

$$AA^{-1}= A^{-1}A=I$$, where $$I$$ is  the Identity matrix.

## Step-by-step guide to finding the inverse of $$2×2$$ matrix

The inverse calculation of a $$2×2$$ matrix is easier compared to higher-order matrices. We can calculate the inverse of a $$2×2$$ matrix using the general steps of calculating the inverse of a matrix.

Let’s find the inverse of the $$2×2$$ matrices below:

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

$$\color{blue}{A^{-1}=(\frac{1}{|A|})\times Adj A=[\frac{1}{(ad-bc)}]\times \begin{bmatrix}d & -b \\-c & a \end{bmatrix}}$$

Note: $$A^{-1}$$ exists only when $$ad-bc≠ 0$$.

Therefore, to calculate the inverse of the $$2×2$$ matrix, we must first change the positions of the expressions $$a$$ and $$d$$, put a negative sign for the expressions $$b$$, and $$c$$, and finally divide it by the determinant of the matrix.

### Inverse of $$2×2$$ Matrix – Example 1:

Find the inverse of matrix $$A$$. $$A= \begin{bmatrix}5 & -1 \\6 & 2 \end{bmatrix}$$

First, find the determinant $$A$$: $$|A|=(5)(2)-(-1)(6)=10-(-6)=10+6=16$$

Then, use this formula: $$\color{blue}{A^{-1}=[\frac{1}{(ad-bc)}]\times \begin{bmatrix}d & -b \\-c & a \end{bmatrix}}$$

$$A^{-1}=\frac{1}{16} \begin{bmatrix}2 & 1 \\-6 & 5 \end{bmatrix}$$

$$=\begin{bmatrix}\frac{1}{16} (2)& \frac{1}{16}(1)\\\frac{1}{16}(-6)& \frac{1}{16}(5)\end{bmatrix}$$

$$=\begin{bmatrix}\frac{2}{16} & \frac{1}{16}\\-\frac{6}{16}& \frac{5}{16}\end{bmatrix}$$

$$A^{-1}= \begin{bmatrix}\frac{1}{8} & \frac{1}{16}\\-\frac{3}{8}& \frac{5}{16}\end{bmatrix}$$

### Inverse of $$2×2$$ Matrix – Example 2:

Find the inverse of matrix $$A$$. $$A= \begin{bmatrix}3 & 5 \\0 & 9 \end{bmatrix}$$

First, find the determinant $$A$$: $$|A|=(3)(9)-(5)(0)=27-0=27$$

Then, use this formula: $$\color{blue}{A^{-1}=[\frac{1}{(ad-bc)}]\times \begin{bmatrix}d & -b \\-c & a \end{bmatrix}}$$

$$A^{-1}=\frac{1}{27} \begin{bmatrix}9 & -5 \\-0 & 3 \end{bmatrix}$$

$$=\begin{bmatrix}\frac{1}{27} (9)& \frac{1}{27}(-5)\\\frac{1}{27}(-0)& \frac{1}{27}(3)\end{bmatrix}$$

$$=\begin{bmatrix}\frac{9}{27} & -\frac{5}{27}\\-\frac{0}{27}& \frac{3}{27}\end{bmatrix}$$

$$A^{-1}= \begin{bmatrix}\frac{1}{3} & -\frac{5}{27}\\ 0& \frac{1}{9}\end{bmatrix}$$

## Exercise for Inverse of $$2×2$$ Matrix

### Find the inverse of each matrix.

1. $$\color{blue}{\begin{bmatrix}5 & -2 \\-1 & 2 \end{bmatrix}}$$
2. $$\color{blue}{\begin{bmatrix}4 & 6 \\2 & -4 \end{bmatrix}}$$
3. $$\color{blue}{\begin{bmatrix}3 & 4 \\6 & 8 \end{bmatrix}}$$
4. $$\color{blue}{\begin{bmatrix}5 & 2 \\-7& -3 \end{bmatrix}}$$
5. $$\color{blue}{\begin{bmatrix}-8 & 3 \\5& 9 \end{bmatrix}}$$
1. $$\color{blue}{\begin{bmatrix}\frac{1}{4} & \frac{1}{4} \\ \frac{1}{8} & \frac{5}{8} \end{bmatrix}}$$
2. $$\color{blue}{\begin{bmatrix}\frac{1}{7} & \frac{3}{14} \\ \frac{1}{14} & -\frac{1}{7} \end{bmatrix}}$$
3. $$\color{blue}{ matrix\: is\: singular}$$
4. $$\color{blue}{\begin{bmatrix} 3 & 2 \\ -7 & -5 \end{bmatrix}}$$
5. $$\color{blue}{\begin{bmatrix}-\frac{3}{29} & \frac{1}{29} \\ \frac{5}{87} & \frac{8}{87} \end{bmatrix}}$$

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