JEE Advance - Mathematics (2023 - Paper 2 Online - No. 5)
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?
$M$ is invertible
There exists a nonzero column matrix $\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)$ such that $M\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)=\left(\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right)$
The set $\left\{X \in \mathbb{R}^3: M X=\mathbf{0}\right\} \neq\{\mathbf{0}\}$, where $\mathbf{0}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)$
The matrix $(M-2 I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
Explanation
$M=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$
$|M|=-1+1=0 \Rightarrow M$ is singular so non-invertible $$ \Rightarrow $$ [A] is wrong.
(B) $M\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right] \Rightarrow\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right]$
$$ \left.\begin{array}{l} a_1+a_2+a_3=-a_1 \\ a_1+a_3=-a_2 \\ a_2=-a_3 \end{array}\right\} \Rightarrow a_1=0 \text { and } a_2+a_3=0 \text { infinite solutions exists [B] is correct. } $$
Option (D) :
$$ \begin{aligned} & M-2 I=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]-2\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{array}\right] \\\\ & |M-2 I|=0 \Rightarrow[D] \text { is wrong } \end{aligned} $$
Option (C) :
$$ \begin{aligned} & \mathrm{MX}=0 \Rightarrow\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\\\ & x+y+z=0 \\\\ & x+z=0 \\\\ & y=0 \\\\ & \therefore \text { Infinite solution } \end{aligned} $$
$[\mathrm{C}]$ is correct
$|M|=-1+1=0 \Rightarrow M$ is singular so non-invertible $$ \Rightarrow $$ [A] is wrong.
(B) $M\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right] \Rightarrow\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right]$
$$ \left.\begin{array}{l} a_1+a_2+a_3=-a_1 \\ a_1+a_3=-a_2 \\ a_2=-a_3 \end{array}\right\} \Rightarrow a_1=0 \text { and } a_2+a_3=0 \text { infinite solutions exists [B] is correct. } $$
Option (D) :
$$ \begin{aligned} & M-2 I=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]-2\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{array}\right] \\\\ & |M-2 I|=0 \Rightarrow[D] \text { is wrong } \end{aligned} $$
Option (C) :
$$ \begin{aligned} & \mathrm{MX}=0 \Rightarrow\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\\\ & x+y+z=0 \\\\ & x+z=0 \\\\ & y=0 \\\\ & \therefore \text { Infinite solution } \end{aligned} $$
$[\mathrm{C}]$ is correct
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