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(A) The given system of equations is $x+y+z=6$,$5x-y+2z=3$,and $2x+y-z=-5$.In matrix form $AX=D$,where $A=\left[\begin{array}{ccc}1 & 1 & 1 \ 5 & -1

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$\left[\begin{array}{c}-15 \ 30 \ 75\end{array}ight]$

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(A) The given system of equations is $x+y+z=6$,$5x-y+2z=3$,and $2x+y-z=-5$.In matrix form $AX=D$,where $A=\left[\begin{array}{ccc}1 & 1 & 1 \ 5 & -1 & 2 \ 2 & 1 & -1\end{array}ight]$,$X=\left[\begin{array}{c}x \ y \ z\end{array}ight]$,and $D=\left[\begin{array}{c}6 \ 3 \ -5\end{array}ight]$.First,we find the cofactor matrix $C$ of $A$:$C_{11} = (-1)^{1+1} \begin{vmatrix} -1 & 2 \ 1 & -1 \end{vmatrix} = 1-2 = -1$$C_{12} = (-1)^{1+2} \begin{vmatrix} 5 & 2 \ 2 & -1 \end{vmatrix} = -(-5-4) = 9$$C_{13} = (-1)^{1+3} \begin{vmatrix} 5 & -1 \ 2 & 1 \end{vmatrix} = 5-(-2) = 7$$C_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 1 \ 1 & -1 \end{vmatrix} = -(-1-1) = 2$$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \ 2 & -1 \end{vmatrix} = -1-2 = -3$$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 1 \ 2 & 1 \end{vmatrix} = -(1-2) = 1$$C_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 1 \ -1 & 2 \end{vmatrix} = 2-(-1) = 3$$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 1 \ 5 & 2 \end{vmatrix} = -(2-5) = 3$$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 1 \ 5 & -1 \end{vmatrix} = -1-5 = -6$Thus,$\operatorname{adj}(A) = C^T = \left[\begin{array}{ccc}-1 & 2 & 3 \ 9 & -3 & 3 \ 7 & 1 & -6\end{array}ight]$.Now,$(\operatorname{adj} A)D = \left[\begin{array}{ccc}-1 & 2 & 3 \ 9 & -3 & 3 \ 7 & 1 & -6\end{array}ight] \left[\begin{array}{c}6 \ 3 \ -5\end{array}ight] = \left[\begin{array}{c}-6+6-15 \ 54-9-15 \ 42+3+30\end{array}ight] = \left[\begin{array}{c}-15 \ 30 \ 75\end{array}ight]$.

If $AX=D$ represents the system of simultaneous linear equations $x+y+z=6$,$5x-y+2z=3$ and $2x+y-z=-5$,then $(\operatorname{Adj} A)D=$

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