Solve this
Question:

$5 x-7 y+z=11$

$6 x-8 y-z=15$

$3 x+2 y-6 z=7$

Solution:

Given $5 x-7 x+7=11$

$6 x-8 y-z=15$

$3 x+2 y-6 z=7$

$D=\left|\begin{array}{ccc}5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & -6\end{array}\right|$

$=5(48+2)+7(-36+3)+1(12+24)$

$=5(50)+7(-33)+1(36)$

$=55$

$D_{1}=\left|\begin{array}{ccc}11 & -7 & 1 \\ 15 & -8 & -1 \\ 7 & 2 & -6\end{array}\right|$

$=11(48+2)+7(-90+7)+1(30+56)$

$=11(50)+7(-83)+1(86)$

$=55$

$D_{2}=\left|\begin{array}{ccc}5 & 11 & 1 \\ 6 & 15 & -1 \\ 3 & 7 & -6\end{array}\right|$

$=5(-90+7)-11(-36+3)+1(42-45)$

$=5(-83)-11(-33)+1(-3)$

$=-55$

$D_{3}=\left|\begin{array}{ccc}5 & -7 & 11 \\ 6 & -8 & 15 \\ 3 & 2 & 7\end{array}\right|$

$=5(-56-30)+7(42-45)+11(12+24)$

$=5(-86)+7(-3)+11(36)$

$=-55$

Now,

$x=\frac{D_{1}}{D}=\frac{55}{55}=1$

$y=\frac{D_{2}}{D}=\frac{-55}{55}=-1$

$z=\frac{D_{3}}{D}=\frac{-55}{55}=-1$

$\therefore x=1, y=-1$ and $z=-1$

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