Q69 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

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

  • A
    No solution
  • Correct answer
    B
    A unique solution
  • C
    More than one but finite solutions
  • D
    Infinitely many solutions

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.

Hence, the correct answer is option: 'B'.

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