Inverse of a 2×2 matrix

In matrix algebra, we can add, subtract and multiply matrices as long as the matrix order is correct. Unlike traditional arithmetic, we cannot divide matrices. Instead, we multiply by the inverse matrix. Inverse matrices have many applications, including computer animation, encryption and digital image transformations.

Inverse matrices

An inverse matrix is the square matrix of the same order that you multiply by to get an identity matrix.

For example, if \(A\) is a square matrix and \(B\) is another square matrix of the same order such that \(AB=BA=I\), then we call \(B\) the inverse of \(A\).

The inverse of \(A\) is denoted by the symbol \(A^{-1}\).

\[AA^{-1}=A^{-1}A=I\]

Compare this to traditional arithmetic, where an inverse is the reciprocal, or \(1\) divided by the number. In matrix algebra, \(A^{-1}\neq\dfrac{1}{A}\) is not defined.

Not every square matrix has an inverse. If the determinant of a matrix equals \(0\), the inverse does not exist and the matrix is called singular. If the determinant is not \(0\), the inverse exists and we call the matrix non-singular or invertible.

Inverting \(2\times2\) matrices

If \(A\) is a \(2\times2\) matrix, then \(A^{-1}\) is also a \(2\times2\) matrix such that:
\[AA^{-1} = A^{-1}A = \left[ \begin{array}{cc} 1 & 0\\
0 & 1
\end{array} \right] \]

There is a simple formula to find the matrix of a \(2\times2\) matrix.

For: \[A = \left[ \begin{array}{cc} a & b\\
c & d
\end{array} \right] \] the inverse matrix of \(A\), or \(A^{-1}\), is given by: \[A^{-1} = \frac{1}{ad-bc}\left[ \begin{array}{cc} d & -b\\
-c & a
\end{array} \right] \]

You may notice that \(ad-bc\) is the determinant of the matrix \(A\). We can rewrite the formula using \(\det A\) or \(\left|A\right|\).

\[A^{-1} = \frac{1}{\det A}\left[ \begin{array}{cc} d & -b\\
-c & a
\end{array} \right] \]

If \(\det A=0\), we have:

\[A^{-1} = \frac{1}{0}\left[ \begin{array}{cc} d & -b\\
-c & a
\end{array} \right] \]

But, \(\dfrac{1}{0}\) is undefined, so the inverse does not exist when \(\det A=0\).

Example 1 – finding the inverse of \(2\times2\) matrices

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

First, check if the inverse of \(A\) exists. To do this, calculate the determinant and determine whether \(A\) is singular, i.e. \(\det A=0\).

\[\begin{align*} \det A & = (2\times4)-(2\times3)\\
& = 8-6\\
& = 2
\end{align*}\]

\(\det A\neq0\), so \(A\) is not singular and the inverse exists. We can use the formula to calculate the inverse. \[\begin{align*} A^{-1} & = \frac{1}{\det A}\left[ \begin{array}{cc} 4 & -3\\
-2 & 2
\end{array} \right]\\
& = \frac{1}{2}\left[ \begin{array}{cc} 4 & -3\\
-2 & 2
\end{array} \right]\\
& = \left[\begin{array}{cc} 2 & -\frac{3}{2}\\
-1 & 1
\end{array} \right]
\end{align*}\]

We can check that \(A^{-1}\) satisfies \(AA^{-1}=I\) and \(A^{-1}A=I\).

\[\begin{align*} AA^{-1} & = \left[ \begin{array}{cc} 2 & 3\\
2 & 4
\end{array} \right]
\left[ \begin{array}{cc} 2 & -\frac{3}{2}\\
-1 & 1
\end{array} \right]\\
& = \left[\begin{array}{cc} (2\times2)+(3\times(-1)) & (2\times(-\frac{3}{2}))+(3\times1)\\
(2\times2)+(4\times(-1)) & (2\times(-\frac{3}{2}))+(4\times1)
\end{array} \right]\\
& = \left[ \begin{array}{cc} 1 & 0\\
0 & 1
\end{array} \right]
\end{align*}\] \[\begin{align*} A^{-1}A & = \left[ \begin{array}{cc} 2 & -\frac{3}{2}\\
-1 & 1
\end{array} \right]
\left[ \begin{array}{cc} 2 & 3\\
2 & 4
\end{array} \right]\\
& = \left[ \begin{array}{cc} (2\times2)+((-\frac{3}{2})\times2) & (2\times3)+((-\frac{3}{2})\times4)\\
(-1\times2)+(1\times2) & ((-1)\times3)+(1\times4)
\end{array} \right]\\
& = \left[ \begin{array}{cc} 1 & 0\\
0 & 1
\end{array} \right]
\end{align*}\]

Therefore, \(A^{-1}=\left[ \begin{array}{cc} 2 & -\frac{3}{2}\\
-1 & 1
\end{array} \right]\).