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(A) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight]$ and $B = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 &

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$x=0, y=5, z=3$

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(A) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight]$ and $B = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight]$.First,we calculate the product $AB$:$AB = \left[\begin{array}{ccc}1(-2)+(-1)(9)+2(6) & 1(0)+(-1)(2)+2(1) & 1(1)+(-1)(-3)+2(-2) \ 0(-2)+2(9)+(-3)(6) & 0(0)+2(2)+(-3)(1) & 0(1)+2(-3)+(-3)(-2) \ 3(-2)+(-2)(9)+4(6) & 3(0)+(-2)(2)+4(1) & 3(1)+(-2)(-3)+4(-2)\end{array}ight]$$= \left[\begin{array}{ccc}-2-9+12 & 0-2+2 & 1+3-4 \ 0+18-18 & 0+4-3 & 0-6+6 \ -6-18+24 & 0-4+4 & 3+6-8\end{array}ight] = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] = I$.Since $AB = I$,$A^{-1} = B = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight]$.The system of equations can be written as $AX = C$,where $X = \left[\begin{array}{c}x \ y \ z\end{array}ight]$ and $C = \left[\begin{array}{c}1 \ 1 \ 2\end{array}ight]$.Then $X = A^{-1}C = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight] \left[\begin{array}{c}1 \ 1 \ 2\end{array}ight]$.$X = \left[\begin{array}{c}-2(1)+0(1)+1(2) \ 9(1)+2(1)+(-3)(2) \ 6(1)+1(1)+(-2)(2)\end{array}ight] = \left[\begin{array}{c}-2+0+2 \ 9+2-6 \ 6+1-4\end{array}ight] = \left[\begin{array}{c}0 \ 5 \ 3\end{array}ight]$.Thus,$x=0, y=5, z=3$.

Use the product $\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{lll}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations:
$x-y+2z=1$
$2y-3z=1$
$3x-2y+4z=2$

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Use the product $\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{lll}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations:$x-y+2z=1$$2y-3z=1$$3x-2y+4z=2$