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.

Related Topics

• How to Add and Subtract Matrices
• How to Multiply Matrix
• How to Multiply a Matrix by a Scalar
• How to Find Determinants of a Matrix

A 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. \(\color{blue}{\begin{bmatrix}-\frac{3}{29} & \frac{1}{29} \\ \frac{5}{87} & \frac{8}{87} \end{bmatrix}}\)