Question 15.15: Find the inverse matrix A^−1 for matrix A = [2 4 3 3 5 0 4 2......
Find the inverse matrix A^{-1} for matrix A = \begin{bmatrix} 2 &4 &3 \\ 3 &5 &0\\ 4& 2& 5 \end{bmatrix}
We have already determined the adjoint for this particular matrix in the example above. Its determinant |A| can be evaluated by expanding down the 3rd column as
|A| = 3 \begin{vmatrix} 3 &5\\ 4& 2 \end{vmatrix} ∣ − 0 + 5 \begin{vmatrix} 2 &4\\ 3& 5 \end{vmatrix}
= 3(6 − 20) + 5(10 − 12)
= 3(−14) + 5(−2)
= −42 − 10 = −52
Therefore, given that we already know that AdjA = \begin{bmatrix}25& −14& −15\\ −15& −2& 9 \\−12& 12& −2 \end{bmatrix}
the inverse matrix will be
A^{-1} = \frac{AdjA}{\left|A\right| } = \frac{\begin{bmatrix}25& −14& −15\\ −15& −2& 9 \\−12& 12& −2 \end{bmatrix}}{- 52} = \begin{bmatrix} −0.48 & 0.27& 0.29 \\ 0.29 &0.04& −0.17 \\ 0.27& −0.23& 0.04 \end{bmatrix}