Solve this

Question:

$3 x-y+2 z=3$

$2 x+y+3 z=5$

$x-2 y-z=1$

Solution:

Given: $3 x-y+2 z=3$

$2 x+y+3 z=5$

$x-2 y-z=1$

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

$=3(-1+6)+1(-2-3)+2(-4-1)$

$=0$

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

$=3(-1+6)+1(-5-3)+2(-10-1)$

$=-15$

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

$=3(-5-3)-3(-2-3)+2(2-5)$

$=-15$

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

$=3(1+10)+1(2-5)+3(-4-1)$

$=-15$

Here, $D$ is zero, but $D_{1}, D_{2}$ and $D_{3}$ are non-zero. Thus, the system of linear equations is inconsistent.

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