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Question
If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] ,find A–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and –2y + z = 7.
Solution
\[\text { Here }, \]
\[ A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\]
\[\left| A \right|=1 \left( 1 + 6 \right) + 2\left( 2 - 0 \right) + 0\left( - 4 - 0 \right)\]
\[ = 7 + 4 + 0\]
\[ = 11\]
\[\text{Let C_ij {be the cofactors of the elements a}_ij in A}=\left[ a_ij \right]. \text { Then },\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ - 2 & 1\end{vmatrix} = 7, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 0 & 1\end{vmatrix} = - 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 0 & - 2\end{vmatrix} = - 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 2 & 0 \\ - 2 & 1\end{vmatrix} = 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 2 \\ 0 & - 2\end{vmatrix} = 2\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 2 & 0 \\ 1 & 3\end{vmatrix} = - 6, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 0 \\ 2 & 3\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 2 \\ 2 & 1\end{vmatrix} = 5\]
\[\therefore adj A = \begin{bmatrix}7 & - 2 & - 4 \\ 2 & 1 & 2 \\ - 6 & - 3 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[or, AX = B\]
\[\text { where }, A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} \text { and }B = \begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
\[\text { Now }, \]
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{11}\begin{bmatrix}70 + 16 - 42 \\ - 20 + 8 - 21 \\ - 40 + 16 + 35\end{bmatrix}\]
\[\Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}44 \\ - 33 \\ 11\end{bmatrix}\]
\[ \therefore x = 4, y = - 3 \text { and } z = 1\]
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