If A = begin{bmatrix} 2 & -3 & 5 3 & 2 & -4 | vedclass

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(A) Given $A = \begin{bmatrix} 2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2 \end{bmatrix}$.First,find the determinant $|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2

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$x=1, y=2, z=3$

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(A) Given $A = \begin{bmatrix} 2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2 \end{bmatrix}$.First,find the determinant $|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 
eq 0$.Next,find the cofactor matrix $C = [C_{ij}]$:$C_{11} = 0, C_{12} = 2, C_{13} = 1$$C_{21} = -1, C_{22} = -9, C_{23} = -5$$C_{31} = 2, C_{32} = 23, C_{33} = 13$Thus,$	ext{adj}(A) = C^T = \begin{bmatrix} 0 & -1 & 2 \ 2 & -9 & 23 \ 1 & -5 & 13 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = -1 \begin{bmatrix} 0 & -1 & 2 \ 2 & -9 & 23 \ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \ -2 & 9 & -23 \ -1 & 5 & -13 \end{bmatrix}$.The system of equations is $AX = B$,where $X = \begin{bmatrix} x \ y \ z \end{bmatrix}$ and $B = \begin{bmatrix} 11 \ -5 \ -3 \end{bmatrix}$.$X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \ -2 & 9 & -23 \ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \ -5 \ -3 \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \ -22 - 45 + 69 \ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$.Therefore,$x=1, y=2, z=3$.

If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$,find $A^{-1}$. Using $A^{-1}$,solve the system of equations: $2x - 3y + 5z = 11$,$3x + 2y - 4z = -5$,and $x + y - 2z = -3$.

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