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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
Let $A$ be a $3 \times 3$ real matrix such that $\mathrm{A}\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=2\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right), \mathrm{A}\left(\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right)=4\left(\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right), \mathrm{A}\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=2\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$.  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
  • ✓
    unique solution
  • B
    exactly two solutions
  • C
    no solution
  • D
    infinitely many solutions
✓

Answer

Correct option: A.
unique solution
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 $A\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right] ....(1)$
$\therefore\left[\begin{array}{c}\mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right]$
$\therefore \mathrm{x}_1+\mathrm{z}_1  =2 ..........(2)$
$\mathrm{x}_2+\mathrm{z}_2  =0 ..............(3)$
$\mathrm{x}_3+\mathrm{z}_3 =0 ..............(4)$
Given $A\left[\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l}-4 \\ 0 \\ 4\end{array}\right]$
$\therefore\left[\begin{array}{l}-\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 4\end{array}\right]$
$\Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 ........(5)$
$-\mathrm{x}_2+\mathrm{x}_2=0  ........(6)$
$-\mathrm{x}_3+\mathrm{z}_3=4$
Given $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$
$\therefore\left[\begin{array}{l}\mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3\end{array}\right]=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]$
$ \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 $
$\therefore $ from $(2), (3), (4), (5), (6)$ and $(7)$
$\mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1$
$ \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0$
$ \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{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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