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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 $A \left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right) ; A \left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right)$ and $A \left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)$. If $X =\left( x _{1}, x _{2}, x _{3}\right)^{ T }$ and $I$ is an identity matrix of order $3$ , then the system $( A -2 I ) X =\left(\begin{array}{l}4 \\ 1 \\ 1\end{array}\right)$ has
  • A
    no solution
  • ✓
    infinitely many solutions
  • C
    unique solution
  • D
    exactly two solutions
✓

Answer

Correct option: B.
infinitely many solutions
$A =\left[\begin{array}{lll} a _{1} & b _{1} & c _{1} \\ a _{2} & b _{2} & c _{2} \\ a _{3} & b _{3} & c _{3}\end{array}\right]$
$A \left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l} c _{1} \\ c _{2} \\ c _{3}\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$
$\Rightarrow c _{1}=1, c _{2}=1, c _{3}=2$
$A \left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]=\left[\begin{array}{l} c _{1}+ a _{1} \\ c _{2}+ a _{2} \\ c _{3}+ a _{3}\end{array}\right]=\left[\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right]$
$\Rightarrow a _{1}=-2, a _{2}=-1, a _{3}=-1$
$A \left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l} a _{1}+ b _{1} \\ a _{2}+ b _{2} \\ a _{3}+ b _{3}\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]$
$\Rightarrow b _{1}=3, b _{2}=2, b _{3}=1$
$\Rightarrow A =\left[\begin{array}{lll}-2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2\end{array}\right]$
$\Rightarrow A -2 I =\left[\begin{array}{ccc}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{array}\right]$
$| A -2 I |=0$
Now, $\left[\begin{array}{lll}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{array}\right]\left[\begin{array}{l} x _{1} \\ x _{2} \\ x _{3}\end{array}\right]=\left[\begin{array}{l}4 \\ 1 \\ 1\end{array}\right]$
$-4 x_{1}+3 x_{2}+x_{3}=4 ...... .$ ($1$)
$- x _{1}+ x _{3}=1 \ldots . (2)$
$- x _{1}+ x _{2}=1 \ldots . (2)$
$(1) -[(2)+3(3)]$
$0=0 \Rightarrow$ infinite solutions

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