Let $A=\left[\begin{array}{rrr}2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1\end{array} ight]$ and $D=\left[\begin{array}{l}0 \ 1 \ 0\end{array} ight]$. The system $A X=D$ has

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Let $A=\left[\begin{array}{rrr}2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1\end{array}ight]$ and $D=\left[\begin{array}{l}0 \ 1 \ 0\end{array}ight]$. The system $A X=D$ has

Aicharya · Accepted Answer

$A=\left[\begin{array}{ccc}2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1\end{array}ight], D=\left[\begin{array}{l}0 \ 1 \ 0\end{array}ight]$ Given, $A X=D$ $\left[\begin{array}{ccc}2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & -1\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{l}0 \ 1 \ 0\end{array}ight]$ $2 x-y+3 z=0$ $x+y-z=1$ $0 x+0 y-z=0$ $\Delta=\left|\begin{array}{ccc}2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & -1\end{array}ight|$ $=2(-1+0)+1(-1+0)+3(0)$ $=-2-1$ $\Delta=-3 
eq 0$ $	herefore$ System of equations have unique solution. Hence, option (2) is correct.

Let $A=\left[\begin{array}{rrr}2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & 1\end{array}\right]$ and…

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