## M8 Inverse of a 2x2 matrix

In matrix algebra, we can add, subtract and multiply matrices subject to conditions on the matrix shape (or order). While matrix algebra does not have a division operation, there is multiplication by the inverse matrix. This module discuses the concept of an inverse matrix.

## Definition

Let $$I$$ denote the identity matrix. That is the matrix containing ones on the main diagonal and zeros elsewhere. That is \begin{align*} I_{1}=\left[1\right],\;I_{2} & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right],\;I_{3}=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right],\;I_{4}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right],\;I_{n}=\left[\begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right]. \end{align*}

If $$A$$ is a square matrix and $$B$$ is another square matrix of the same order such that \begin{align*} AB & =BA=I \end{align*} then we call $$B$$ the inverse of $$A.$$ The inverse of $$A$$ is denoted by the symbol $$A^{-1}$$.1 Note that \begin{align*} A^{-1} & \neq\frac{1}{A} \end{align*} as division is not defined in matrix algebra. Hence \begin{align*} AA^{-1} & =A^{-1}A=I. \end{align*}

Not every square matrix has an inverse. If the determinant of a matrix equals zero, the inverse does not exist and the matrix is called singular. If the determinant is unequal to zero the inverse exists and we call the matrix non-singular or invertible.

## Inverse of a $$2\times2$$ Matrix

If $$A$$ is a $$2\times2$$ matrix, then $$A^{-1}$$ is also a $$2\times2$$ matrix such that: \begin{align*} AA^{-1} & =A^{-1}A=\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]. \end{align*} There is a simple formula to find the matrix of a $$2\times2$$ matrix.

Let \begin{align*} A=\left[\begin{array}{cc} a & b\\ c & d \end{array}\right] \end{align*} then the inverse matrix of $$A$$ is given by \begin{align*} A^{-1} & =\frac{1}{ad-bc}\left[\begin{array}{cc} d & -b\\ -c & a \end{array}\right]. \end{align*}

Note that $$ad-bc$$ is the determinant of the matrix $$A.$$ That is $$ad-bc=\det A=\left|A\right|$$. So we can also write \begin{align*} A^{-1} & =\frac{1}{\det A}\left[\begin{array}{cc} d & -b\\ -c & a \end{array}\right]. \end{align*} If $$\det A=0,$$ we have \begin{align*} A^{-1} & =\frac{1}{0}\left[\begin{array}{cc} d & -b\\ -c & a \end{array}\right]. \end{align*} But $$1/0$$ is undefined and so the inverse does not exist.

#### Example 1

Find the inverse of the matrix $$A=\left[\begin{array}{cc} 2 & 3\\ 2 & 4 \end{array}\right]$$.

Solution:

First check if $$A$$ is singular. \begin{align*} \det A & =2\times4-2\times3\\ & =8-6\\ & =2 \end{align*} so $$A$$ is not singular and the inverse exists. Using the formula above, \begin{align*} A^{-1} & =\frac{1}{\det A}\left[\begin{array}{cc} 4 & -3\\ -2 & 2 \end{array}\right]\\ & =\frac{1}{2}\left[\begin{array}{cc} 4 & -3\\ -2 & 2 \end{array}\right]\\ & =\left[\begin{array}{cc} 2 & -3/2\\ -1 & 1 \end{array}\right] \end{align*} Check $$AA^{-1}=I$$ and $$A^{-1}A=I.$$ \begin{align*} AA^{-1} & =\left[\begin{array}{cc} 2 & 3\\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} 2 & -3/2\\ -1 & 1 \end{array}\right]\\ & =\left[\begin{array}{cc} 2\times2+3\times\left(-1\right) & 2\times\left(-3/2\right)+3\times1\\ 2\times2+4\times\left(-1\right) & 2\times\left(-3/2\right)+4\times1 \end{array}\right]\\ & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]\\ A^{-1}A & =\left[\begin{array}{cc} 2 & -3/2\\ -1 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3\\ 2 & 4 \end{array}\right]\\ & =\left[\begin{array}{cc} 2\times2+\left(-3/2\right)\times2 & 2\times3+\left(-3/2\right)\times4\\ -1\times2+1\times2 & \left(-1\right)\times3+1\times4 \end{array}\right]\\ & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]. \end{align*} So $$A^{-1}=\left[\begin{array}{cc} 2 & -3/2\\ -1 & 1 \end{array}\right]$$ is the inverse of $$A.$$

#### Example 2

Find the inverse of the matrix $$A=\left[\begin{array}{cc} -1 & -2\\ 4 & 3 \end{array}\right]$$.

Solution:

The determinant, \begin{align*} \det A & =-1\times3-\left(-2\right)\times4\\ & =-3-\left(-8\right)\\ & =5 \end{align*} so the matrix $$A$$ has an inverse. Using the formula, \begin{align*} A^{-1} & =\frac{1}{\det A}\left[\begin{array}{cc} 3 & 2\\ -4 & -1 \end{array}\right]\\ & =\frac{1}{5}\left[\begin{array}{cc} 3 & 2\\ -4 & -1 \end{array}\right]. \end{align*} Check $$AA^{-1}=I$$ and $$A^{-1}A=I.$$ \begin{align*} AA^{-1} & =\left[\begin{array}{cc} -1 & -2\\ 4 & 3 \end{array}\right]\left(\frac{1}{5}\left[\begin{array}{cc} 3 & 2\\ -4 & -1 \end{array}\right]\right)\\ & =\frac{1}{5}\left[\begin{array}{cc} -1 & -2\\ 4 & 3 \end{array}\right]\left[\begin{array}{cc} 3 & 2\\ -4 & -1 \end{array}\right]\\ & =\frac{1}{5}\left[\begin{array}{cc} -3+8 & -2+2\\ 12-12 & 8-3 \end{array}\right]\\ & =\frac{1}{5}\left[\begin{array}{cc} 5 & 0\\ 0 & 5 \end{array}\right]\\ & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]\\ A^{-1}A & =\frac{1}{5}\left[\begin{array}{cc} 3 & 2\\ -4 & -1 \end{array}\right]\left[\begin{array}{cc} -1 & -2\\ 4 & 3 \end{array}\right]\\ & =\frac{1}{5}\left[\begin{array}{cc} -3+8 & -6+6\\ 4-4 & 8-3 \end{array}\right]\\ & =\frac{1}{5}\left[\begin{array}{cc} 5 & 0\\ 0 & 5 \end{array}\right]\\ & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]. \end{align*} So $$A^{-1}=\frac{1}{5}\left[\begin{array}{cc} 3 & 2\\ -4 & -1 \end{array}\right].$$

#### Example 3

Find the inverse of the matrix $$A=\left[\begin{array}{cc} 3 & 2\\ 6 & 4 \end{array}\right].$$

Solution: The determinant of $$A$$ is \begin{align*} \det A & =3\times4-6\times2\\ & =12-12\\ & =0. \end{align*} Since $$\det A=0,$$the matrix $$A$$ does not have an inverse.

## Exercise 1

Find if possible, the inverses of the following matrices:

\begin{align*} a)\, & \left[\begin{array}{cc} 2 & -1\\ -4 & 3 \end{array}\right] & b) & \left[\begin{array}{cc} 0 & 4\\ 2 & 5 \end{array}\right] & c) & \left[\begin{array}{cc} -2 & -3\\ 6 & 9 \end{array}\right] & d) & \left[\begin{array}{cc} -3 & 4\\ 2 & 1 \end{array}\right]. \end{align*}

\begin{align*} \begin{array}{llll} a)\,\left[\begin{array}{cc} 3/2 & 1/2\\ 2 & 1 \end{array}\right] & b)\,\left[\begin{array}{cc} -5/8 & 1/2\\ 1/4 & 0 \end{array}\right] & c)\,\textrm{Inverse does not exist} & d)\,\frac{1}{11}\left[\begin{array}{cc} -1 & 4\\ 2 & 3 \end{array}\right].\end{array} \end{align*}

## Exercise 2

For

\begin{align*} A & =\left[\begin{array}{cc} -1 & -3\\ 2 & 5 \end{array}\right]\textrm{ and B=\left[\begin{array}{cc} -2 & 0\\ 4 & 1 \end{array}\right] } \end{align*} find $$A^{-1}$$ and $$B^{-1}$$ and show that $$\left(AB\right)^{-1}=B^{-1}A^{-1}.$$

\begin{align*} A^{-1} & =\left[\begin{array}{cc} 5 & 3\\ -2 & -1 \end{array}\right],\quad B^{-1}=\left[\begin{array}{cc} -1/2 & 0\\ 2 & 1 \end{array}\right]. \end{align*}