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JEE Main 31-Jan-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 31-Jan-2024 Paper - Shift 24 Marks
MCQ
Let $A$ be a $33$ real matrix such that $ A\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right)=4\left(\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right), A\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=2\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)$
Then, the system $(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ has
  • A
    unique solution
  • B
    exactly two solutions
  • C
    no solution
  • D
    infinitely many solutions
✓

Answer

Let $A =\left[\begin{array}{lll} x _1 & y _1 & z _1 \\ x _2 & y _2 & z _2 \\ x _3 & y _3 & z _3\end{array}\right]$
Given $A\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right]$
$\therefore\left[\begin{array}{l} x _1+ z _1 \\ x _2+ z _2 \\ x _3+ z _3\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right]$
$\therefore x_1+z_1 =2$
$x_2+z_2 =0$
$x_3+z_3 =0$
Given $A\left[\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l}-4 \\ 0 \\ 4\end{array}\right]$
$\therefore\left[\begin{array}{l}- x _1+ z _1 \\ - x _2+ z _2 \\ - x _3+ z _3\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 4\end{array}\right]$
$\Rightarrow-x_1+z_1=-4$
$-x_2+x_2=0$
$-x_3+z_3=4$
Given $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$
$\therefore\left[\begin{array}{l} y _1 \\ y _2 \\ y _3\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$
$\therefore y_1=0, y_2=2, y_3=0$
$\therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7)$
$x_1=3 x, x_2=0, x_3=-1$
$y_1=0, y_2=2, y_3=0$
$z_1=-1, z_2=0, z_3=3$
$\therefore A=\left[\begin{array}{ccc}3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3\end{array}\right]$
$\therefore \operatorname{Now}(A-31)\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}-1 \\ 2 \\ 3\end{array}\right]$
$\therefore\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]$
$\left[\begin{array}{l}- z \\ - y \\ - x \end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
$[z=-1],[y=-2],[x=-3]$

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