Question 16.6.2: Find the inverse of the following matrix, when it exists: A ......
Find the inverse of the following matrix, when it exists:
A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}We find a 2 ×2 matrix X such that AX = I, after which it is easy to check that XA = I. Solving AX = I requires finding numbers x, y, z, and w such that
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x & y \\ z & w \end{pmatrix} =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}Matrix multiplication implies that
ax + bz = 1, cx + dz = 0, ay + bw = 0, and cy + dw = 1
Note that we have two different systems of equations here. One is given by the two equations on the left, and the other by the two equations on the right. Both these systems have A as a common coefficient matrix. If |A| = ad − bc \neq 0, solving the two pairs of simultaneous equations separately, using Cramer’s rule from Section 16.1, yields
x_{1}=\frac{1}{\left|A\right| } \begin{vmatrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{vmatrix} \ \ \ \text{and} \ \ \ x_{2}=\frac{1}{\left|A\right| } \begin{vmatrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{vmatrix} Cramer’s rule
x={\frac{d}{a d-b c}},\;z={\frac{-c}{a d-b c}},y=\frac{-b}{ad-bc},\;\mathrm{and}\;w={\frac{a}{a d-b c}}Hence, we have proved the following result:
INVERSE OF A MATRIX OF ORDER 2
Provided that |A| = ad − bc \neq 0,
A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\Rightarrow A^{-1}=\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} (16.6.3)
Note that in the inverse matrix, the diagonal elements of the original 2 ×2 matrix are switched, whereas the off-diagonal elements just change sign.