written 4.6 years ago by |
Solution:
let A = \begin{bmatrix} 1 & 3 & 2\ 3 & -2 & 5\ 2 & -3 & 6 \end{bmatrix}, B = \begin{bmatrix} 6 \ 5 \ 7 \end{bmatrix} C = \begin{bmatrix} x \ y \ z \end{bmatrix}
|A| = \begin{vmatrix} 1 & 3 & 2 \ 3 & -2 & 5\ 3 & -3 & 6 \end{vmatrix}
$|A| = 1(-12+15)-3(18-10)+2(-9+4)$
$|A| = -31$
$|A| = 2 \neq 0$
$A^{-1} exists$
$Matrix of minors = \quad \begin{bmatrix} \quad \begin{vmatrix} -2 & 5 \\ -3 & 6 \end{vmatrix} & \quad \begin{vmatrix} 3 & 5 \\ 2 & 6 \end{vmatrix} & \quad \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} \\ \quad \begin{vmatrix} 3 & 2 \\ -3 & 6 \end{vmatrix} & \quad \begin{vmatrix} 1 & 2 \\ 2 & 6 \end{vmatrix} & \quad \begin{vmatrix} 1 & 3 \\ 2 & -3 \end{vmatrix}\\ \quad \begin{vmatrix} 3 & 2 \\ -2 & 5 \end{vmatrix} & \quad \begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix} & \quad \begin{vmatrix} 1 & 3 \\ 3 & -2 \end{vmatrix} \\ \end{bmatrix} $
= \begin{bmatrix} 3 & 8 & -5 \ 24 & 2 & -9\ 19 & -1 & -11 \end{bmatrix}
$Matrix of cofactors = \quad \begin{bmatrix} 3 & -8 & -5 \\ -24 & 2 & 9\\ 19 & 1 & -11 \end{bmatrix}$