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Let $M=\left(a_{i j}\right), i, j \in\{1,2,3\}$, be the $3 \times 3$ matrix such that $a_{i j}=1$ if $j+1$ is divisible by $i$, otherwise $a_{i j}=0$. Then which of the following statements is (are) true?
($A$) $\mathrm{M}$ is invertible
($B$) There exists a nonzero column matrix $\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)$ such that $M\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)=\left(\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right)$
($C$) The set $\left\{\mathrm{X} \in \mathbb{R}^3: \mathrm{MX}=\mathbf{0} \neq \neq 0\right\}$, where $\mathbf{0}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)$
($D$) The matrix $(M-2 I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
$B,C$
$B,D$
$B,A$
$A,C,D$
Solution
$M=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$
$|M|=-1+1=0 \Rightarrow M$ is singular so non-invertible
(B) $\mathrm{M}\left[\begin{array}{l}\mathrm{a}_1 \\ \mathrm{a}_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right] \Rightarrow\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right]$
$\left.\begin{array}{l}a_1+a_2+a_3=-a_1 \\ a_1+a_3=-a_2 \\ a_2=-a_3\end{array}\right\} \Rightarrow a_1=0$ and $a_2+a_3=0$ infinite solutions exists [B] is correct.
Option $(D)$
$\mathrm{M}-2 \mathrm{I}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]-2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2\end{array}\right]$
$|M-2 I|=0 \Rightarrow[D]$ is wrong Option ($C$) :
$\mathrm{MX}=0 \Rightarrow\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
$x+y+z=0$
$x+z=0$
$y=0$
$\therefore$ Infinite solution
$[\mathrm{C}]$ is correct
Similar Questions
Match the Statements / Expressions in Column $I$ with the Statements / Expressions in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS$.
| Column $I$ | Column $II$ |
| $(A)$ The minimum value of $\frac{x^2+2 x+4}{x+2}$ is | $(p)$ $0$ |
| $(B)$ Let $A$ and $B$ be $3 \times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skewsymmetric, and $(A+B)(A-B)=(A-B)(A+B)$. If $(A B)^t=(-1)^k A B$, where $(A B)^t$ is the transpose of the matrix $A B$, then the possible values of $k$ are | $(q)$ $1$ |
| $(C)$ Let $\mathrm{a}=\log _3 \log _3 2$. An integer $\mathrm{k}$ satisfying $1<2^{\left(-k+3^{-2}\right)}<2$, must be less than | $(r)$ $2$ |
| $(D)$ If $\sin \theta=\cos \phi$, then the possible values of $\frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right)$ are | $(s)$ $3$ |
