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Let $A=\left[\begin{array}{rrr}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & 1\end{array}\right]$ and $D=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$. The system $A X=D$ has
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A unique solution
$A=\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & 1\end{array}\right], D=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$
Given, $A X=D$
$\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$
$2 x-y+3 z=0$
$x+y-z=1$
$0 x+0 y-z=0$
$\Delta=\left|\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & -1\end{array}\right|$
$=2(-1+0)+1(-1+0)+3(0)$
$=-2-1$
$\Delta=-3 \neq 0$
$\therefore$ System of equations have unique solution.
Hence, option (2) is correct.
Given, $A X=D$
$\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$
$2 x-y+3 z=0$
$x+y-z=1$
$0 x+0 y-z=0$
$\Delta=\left|\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & -1\end{array}\right|$
$=2(-1+0)+1(-1+0)+3(0)$
$=-2-1$
$\Delta=-3 \neq 0$
$\therefore$ System of equations have unique solution.
Hence, option (2) is correct.
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