Solve this

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Question:

$2 y-3 z=0$

$x+3 y=-4$

$3 x+4 y=3$

Solution:

These equations can be written as

$0 x+2 y-3 z=0$

$x+3 y+0 z=-4$

$3 x+4 y+0 z=3$

$D=\left|\begin{array}{ccc}0 & 2 & -3 \\ 1 & 3 & 0 \\ 3 & 4 & 0\end{array}\right|$

$=0(0-0)-2(0-0)-3(4-9)$

$=15$

$D_{1}=\left|\begin{array}{ccc}0 & 2 & -3 \\ -4 & 3 & 0 \\ 3 & 4 & 0\end{array}\right|$

$=0(0-0)-2(0-0)-3(-16-9)$

$=75$

$D_{2}=\left|\begin{array}{ccc}0 & 0 & -3 \\ 1 & -4 & 0 \\ 3 & 3 & 0\end{array}\right|$

$=0(0-0)-0(0-0)-3(3+12)$

$=-45$

$D_{3}=\left|\begin{array}{ccc}0 & 2 & 0 \\ 1 & 3 & -4 \\ 3 & 4 & 3\end{array}\right|$

$=0(9+16)-2(3+12)-0(4-9)$

$=-30$

Now,

$x=\frac{D_{1}}{D}=\frac{75}{15}=5$

$y=\frac{D_{2}}{D}=\frac{-45}{15}=-3$

$z=\frac{D_{3}}{D}=\frac{-30}{15}=-2$

$\therefore x=5, y=-3$ and $z=-2$

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