Question 3.6.11: Apply the formula in (23) to find the inverse of the matrix ...
Apply the formula in (23) to find the inverse of the matrix
\textbf{A} = \left [ \begin{matrix} 1 & 4 & 5 \\ 4 & 2 & 5 \\ -3 & 3 & -1 \end{matrix} \right ]
\textbf{A}^{-1} = \frac{[A_{ij}]^{T}}{|\textbf{A}|}, (23)
of Example 10, in which we saw that |A| = 29.
First we calculate the cofactors of A, arranging our computations in a natural 3 × 3 array:
A_{11}= + \begin{vmatrix} 2 & 5 \\ 3 & -1 \end{vmatrix} = -17 , A_{12}= – \begin{vmatrix} 4 & 5 \\ -3 & -1 \end{vmatrix}=-11 , A_{13}= +\begin{vmatrix} 4 &2 \\ -3& 3\end{vmatrix} = 18,A_{21}= – \begin{vmatrix} 4 & 5 \\ 3 & -1 \end{vmatrix} = 19 , A_{22}= + \begin{vmatrix} 1 & 5 \\ -3 & -1 \end{vmatrix} = 14 , A_{23}= – \begin{vmatrix} 1 & 4 \\ -3 & 3 \end{vmatrix} = -15 ,
A_{31}= + \begin{vmatrix} 4 & 5 \\ 2 & 5 \end{vmatrix} = 10 , A_{32}= – \begin{vmatrix} 1 & 5 \\ 4 & 5 \end{vmatrix} = 15 , A_{33}= + \begin{vmatrix} 1 & 4 \\ 4 & 2 \end{vmatrix} = -14 .
(Note the familiar checkerboard pattern of signs.) Thus the cofactor matrix of A is
[A_{ij}] = \left [ \begin{matrix} -17 & -11 & 18 \\ 19 & 14 & -15 \\ 10 & 15 & -14 \end{matrix} \right ].
Next, we interchange rows and columns to obtain the adjoint matrix
adj A =[A_{ij}]^{T} = \left [ \begin{matrix} -17 & 19 & 10 \\ -11 & 14 & -15 \\ 18 & -15 & -14 \end{matrix} \right ].
Finally, in accord with Eq. (23), we divide by |A| = 29 to get the inverse matrix
A^{-1} = \frac{1}{29} \left [ \begin{matrix} -17 & 19 & 10 \\ -11 & 14 & -15 \\ 18 & -15 & -14 \end{matrix} \right ].