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