Apply the formula in (23) to find the inverse of the matrix A =[1 4 5 4 2 5 -3 3 -1] of Example 10, in which we saw that |A| = 29.

Question

Apply the formula in (23) to find the inverse of the matrix

[latex] extbf{A} = \left [ \begin{matrix} 1 & 4 & 5 \ 4 & 2 & 5 \ -3 & 3 & -1 \end{matrix} ight ][/latex]

[latex] extbf{A}^{-1} = \frac{[A_{ij}]^{T}}{| extbf{A}|},[/latex] (23)

of Example 10, in which we saw that |A| = 29.

Accepted Answer

First we calculate the cofactors of A, arranging our computations in a natural 3 × 3 array:

[latex]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,[/latex]

[latex]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 ,[/latex]

[latex]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 .[/latex]

(Note the familiar checkerboard pattern of signs.) Thus the cofactor matrix of A is

[latex][A_{ij}] = \left [ \begin{matrix} -17 & -11 & 18 \ 19 & 14 & -15 \ 10 & 15 & -14 \end{matrix} ight ][/latex].

Next, we interchange rows and columns to obtain the adjoint matrix

adj A =[latex][A_{ij}]^{T} = \left [ \begin{matrix} -17 & 19 & 10 \ -11 & 14 & -15 \ 18 & -15 & -14 \end{matrix} ight ][/latex].

Finally, in accord with Eq. (23), we divide by |A| = 29 to get the inverse matrix

A[latex]^{-1} = \frac{1}{29} \left [ \begin{matrix} -17 & 19 & 10 \ -11 & 14 & -15 \ 18 & -15 & -14 \end{matrix} ight ][/latex].

Question 3.6.11: Apply the formula in (23) to find the inverse of the matrix ...

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