Solve this

Question:

$x+y=5$

$y+z=3$

$x+z=4$

Solution:

These equations can be written as

$x+y+0 z=5$

$0 x+y+z=3$

$x+0 y+z=4$

$D=\left|\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right|$

$=1(1-0)-1(0-1)+0(0-1)$

$=1(1)-1(-1)+0$

$=2$

$D_{1}=\left|\begin{array}{lll}5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1\end{array}\right|$

$=5(1-0)-1(3-4)+0(0-4)$

$=5(1)-1(-1)$

$=6$

$D_{2}=\left|\begin{array}{lll}1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1\end{array}\right|$

$=1(3-4)-5(0-1)+0(0-4)$

$=1(-1)-5(-1)$

$=4$

$D_{3}=\left|\begin{array}{lll}1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4\end{array}\right|$

$=1(4-0)-1(0-3)+5(0-1)$

$=1(4)-1(-3)+5(-1)$

$=2$

Now,

$x=\frac{D_{1}}{D}=\frac{6}{2}=3$

$y=\frac{D_{2}}{D}=\frac{4}{2}=2$

$z=\frac{D_{3}}{D}=\frac{2}{2}=1$

$\therefore x=3, y=2$ and $z=1$

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