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Using Cramers rule, solve x-y-2z=1; 2x+3y+4z=4; 3x-2y-6z=5

solution:

D = \begin{bmatrix} 1 & -1 & -2\ 2 & 3 & 4 \ 3 & -2 & -6 \end{bmatrix}

= 1(-18+8)+1(-12-12)-2(-4-9)

=-8

$D_{x}$ = \begin{bmatrix} 1 & -1 & -2\ 4 & 3 & 4 \ 5 & -2 & -6 \end{bmatrix}

= 1(-18+8) +1(-24-20) -2(-8-15)

= -8

therefore, $X = \frac{D_{x}}{D} = \frac{-8}{-8} = 1$

$D_{y}$ = \begin{bmatrix} 1 & 1 & -2\ 2 & 4 & 4 \ 3 & 5 & -6 \end{bmatrix}

=1(-24-20) -1(-12-12)-2(10-12)

=-16

therefore, y = $\frac{D_{y}}{D} = \frac{-16}{-8} = 2$

$D_{z}$ = \begin{bmatrix} 1 & -1 & 1\ 2 & 3 & 4 \ 3 & -2 & 5 \end{bmatrix}

=1(15+8) +1(10-12)+1(-4-9)

= 8

therefore, y = $\frac{D_{z}}{D} = \frac{8}{-8} = -1$

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