Question 16.7.1: Show that the matrix A =(2 3 4 4 3 1 1 2 4) has an inverse, ......
Show that the matrix A=\begin{pmatrix} 2 & 3 & 4 \\ 4 & 3 & 1 \\ 1 & 2 &4 \end{pmatrix} has an inverse, and then find that inverse.
Step-by-Step
According to Theorem 16.7.1, A has an inverse if and only if |A| \neq 0. Here we find that |A| = −5, so the inverse exists. The cofactors are
\begin{matrix} C_{11}=\ \ \ \begin{vmatrix} 3 & 1 \\ 2 & 4\end{vmatrix} =10, & C_{12}=-\begin{vmatrix} 4 & 1 \\ 1 & 4\end{vmatrix} =-15, & C_{13}=\begin{vmatrix} 4& 3 \\ 1 & 2\end{vmatrix} =5\end{matrix}
\begin{matrix} C_{21}=-\begin{vmatrix} 3 & 4 \\ 2 & 4\end{vmatrix} =-4, & C_{22}=\begin{vmatrix} 2 & 4 \\ 1 & 4\end{vmatrix} =4, & C_{23}=-\begin{vmatrix} 2& 3 \\ 1 & 2\end{vmatrix} =-1\end{matrix}
\begin{matrix}C_{31}=\begin{vmatrix} 3 & 4 \\ 3 & 1\end{vmatrix} =-9, & C_{32}=-\begin{vmatrix} 2 & 4 \\ 4 & 1\end{vmatrix} =14, & C_{33}=\begin{vmatrix} 2& 3 \\ 4 & 3\end{vmatrix} =-6\end{matrix}
Hence, the inverse of A is
A^{-1}=\frac{1}{\left|A\right| } \left ( \begin{matrix} C_{11} & C_{21} & C_{31} \\ C_{12} &C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{matrix} \right ) =-\frac{1}{5} \left ( \begin{matrix} 10 & -4 & -9 \\ -15 & 4 & 14 \\ 5 & -1 & -6 \end{matrix} \right )One can check the result by showing that AA^{−1} = I.
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