Let M=(a_i j), 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?

There exists a nonzero column matrix (\beginarrayla_1 a_2 a_3\endarray) such that M(\beginarrayla_1 a_2 a_3\endarray)=(\beginarrayl-a_1 -a_2 -a_3\endarray), The set \X \in \mathbbR^3: M X=0\ eq\0\ , where 0=(\beginarrayl0 0 0\endarray)

There exists a nonzero column matrix (\beginarrayla_1 a_2 a_3\endarray) such that M(\beginarrayla_1 a_2 a_3\endarray)=(\beginarrayl-a_1 -a_2 -a_3\endarray)

The set \X \in \mathbbR^3: M X=0\ eq\0\ , where 0=(\beginarrayl0 0 0\endarray)

The matrix (M-2 I) is invertible, where I is the 3 \times 3 identity matrix

Matrices and Determinants PYQ — JEE Advanced 2023 · Shift 2 · Q22 with Solution

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Correct answer: B, C

Construct the matrix $M$

We are given

1,& \text{if } i \mid (j+1),\\ 0,& \text{otherwise.} \end{cases}$$ for $i,j\in\{1,2,3\}$. We compute row-wise. - For $i=1$: since $1$ divides every integer, all entries are $1$. $$a_{11}=a_{12}=a_{13}=1$$ - For $i=2$: check divisibility of $j+1$ by $2$. - $j=1$: $j+1=2$, divisible by $2$ $\Rightarrow a_{21}=1$ - $j=2$: $j+1=3$, not divisible by $2$ $\Rightarrow a_{22}=0$ - $j=3$: $j+1=4$, divisible by $2$ $\Rightarrow a_{23}=1$ - For $i=3$: check divisibility of $j+1$ by $3$. - $j=1$: $2$ not divisible by $3$ $\Rightarrow a_{31}=0$ - $j=2$: $3$ divisible by $3$ $\Rightarrow a_{32}=1$ - $j=3$: $4$ not divisible by $3$ $\Rightarrow a_{33}=0$ Hence, $$M=\begin{pmatrix} 1&1&1\\ 1&0&1\\ 0&1&0 \end{pmatrix}.$$ --- 2. **Check option A: Is $M$ invertible?** Compute the determinant: $$\det(M)=\begin{vmatrix} 1&1&1\\ 1&0&1\\ 0&1&0 \end{vmatrix}.$$ Expanding along the first row, $$\det(M)=1\begin{vmatrix}0&1\\1&0\end{vmatrix}-1\begin{vmatrix}1&1\\0&0\end{vmatrix}+1\begin{vmatrix}1&0\\0&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1, \qquad \begin{vmatrix}1&1\\0&0\end{vmatrix}=0, \qquad \begin{vmatrix}1&0\\0&1\end{vmatrix}=1.$$ So, $$\det(M)=1(-1)-1(0)+1(1)=0.$$ Therefore $M$ is **not invertible**. So, **A is false**. --- 3. **Check option C: Is the null space nontrivial?** Since $\det(M)=0$, the matrix is singular. Therefore the homogeneous system $$MX=\mathbf 0$$ has a nonzero solution. So, $$\{X\in\mathbb R^3: MX=\mathbf 0\}\neq \{\mathbf 0\}.$$ Hence, **C is true**. We can also find one such vector explicitly. Let $$X=\begin{pmatrix}x\\y\\z\end{pmatrix}.$$ Then $$MX=\begin{pmatrix} x+y+z\\ x+z\\ y \end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$ This gives $$y=0,\quad x+z=0.$$ So one nonzero solution is $$X=\begin{pmatrix}1\\0\\-1\end{pmatrix}.$$ --- 4. **Check option B: Does there exist a nonzero vector such that $MX=-X$?** This means $$(M+I)X=0.$$ So we check whether $M+I$ is singular. $$M+I=\begin{pmatrix} 2&1&1\\ 1&1&1\\ 0&1&1 \end{pmatrix}.$$ Compute its determinant: $$\det(M+I)=\begin{vmatrix} 2&1&1\\ 1&1&1\\ 0&1&1 \end{vmatrix}.$$ Expand along the first row: $$\det(M+I)=2\begin{vmatrix}1&1\\1&1\end{vmatrix}-1\begin{vmatrix}1&1\\0&1\end{vmatrix}+1\begin{vmatrix}1&1\\0&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}1&1\\1&1\end{vmatrix}=0, \qquad \begin{vmatrix}1&1\\0&1\end{vmatrix}=1.$$ Thus, $$\det(M+I)=2(0)-1(1)+1(1)=0.$$ Hence $M+I$ is singular, so there exists a nonzero vector $X$ such that $$(M+I)X=0 \iff MX=-X.$$ Therefore, **B is true**. We can also solve explicitly: $$MX=-X$$ with $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$ gives $$\begin{pmatrix} x+y+z\\x+z\\y \end{pmatrix}=\begin{pmatrix}-x\\-y\\-z\end{pmatrix}.$$ So, \begin{align*} x+y+z&=-x \Rightarrow 2x+y+z=0,\\ x+z&=-y \Rightarrow x+y+z=0,\\ y&=-z \Rightarrow y+z=0. \end{align*} From $y=-z$, and $x+y+z=0$, we get $x=0$. Then $2x+y+z=0$ is automatically satisfied. So one nonzero solution is $$X=\begin{pmatrix}0\\1\\-1\end{pmatrix}.$$ --- 5. **Check option D: Is $(M-2I)$ invertible?** $$M-2I=\begin{pmatrix} -1&1&1\\ 1&-2&1\\ 0&1&-2 \end{pmatrix}.$$ Compute its determinant: $$\det(M-2I)=\begin{vmatrix} -1&1&1\\ 1&-2&1\\ 0&1&-2 \end{vmatrix}.$$ Expand along the first row: \begin{align*} \det(M-2I) &=(-1)\begin{vmatrix}-2&1\\1&-2\end{vmatrix}-1\begin{vmatrix}1&1\\0&-2\end{vmatrix}+1\begin{vmatrix}1&-2\\0&1\end{vmatrix}\\ &=(-1)(4-1)-1(-2)+1(1)\\ &=-3+2+1=0. \end{align*} So $(M-2I)$ is **not invertible**. Hence, **D is false**. --- 6. **Final conclusion** The true statements are: $$\boxed{B,\ C}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: B, C

Construct the matrix $M$

We are given

1,& \text{if } i \mid (j+1),\\ 0,& \text{otherwise.} \end{cases}$$ for $i,j\in\{1,2,3\}$. We compute row-wise. - For $i=1$: since $1$ divides every integer, all entries are $1$. $$a_{11}=a_{12}=a_{13}=1$$ - For $i=2$: check divisibility of $j+1$ by $2$. - $j=1$: $j+1=2$, divisible by $2$ $\Rightarrow a_{21}=1$ - $j=2$: $j+1=3$, not divisible by $2$ $\Rightarrow a_{22}=0$ - $j=3$: $j+1=4$, divisible by $2$ $\Rightarrow a_{23}=1$ - For $i=3$: check divisibility of $j+1$ by $3$. - $j=1$: $2$ not divisible by $3$ $\Rightarrow a_{31}=0$ - $j=2$: $3$ divisible by $3$ $\Rightarrow a_{32}=1$ - $j=3$: $4$ not divisible by $3$ $\Rightarrow a_{33}=0$ Hence, $$M=\begin{pmatrix} 1&1&1\\ 1&0&1\\ 0&1&0 \end{pmatrix}.$$ --- 2. **Check option A: Is $M$ invertible?** Compute the determinant: $$\det(M)=\begin{vmatrix} 1&1&1\\ 1&0&1\\ 0&1&0 \end{vmatrix}.$$ Expanding along the first row, $$\det(M)=1\begin{vmatrix}0&1\\1&0\end{vmatrix}-1\begin{vmatrix}1&1\\0&0\end{vmatrix}+1\begin{vmatrix}1&0\\0&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1, \qquad \begin{vmatrix}1&1\\0&0\end{vmatrix}=0, \qquad \begin{vmatrix}1&0\\0&1\end{vmatrix}=1.$$ So, $$\det(M)=1(-1)-1(0)+1(1)=0.$$ Therefore $M$ is **not invertible**. So, **A is false**. --- 3. **Check option C: Is the null space nontrivial?** Since $\det(M)=0$, the matrix is singular. Therefore the homogeneous system $$MX=\mathbf 0$$ has a nonzero solution. So, $$\{X\in\mathbb R^3: MX=\mathbf 0\}\neq \{\mathbf 0\}.$$ Hence, **C is true**. We can also find one such vector explicitly. Let $$X=\begin{pmatrix}x\\y\\z\end{pmatrix}.$$ Then $$MX=\begin{pmatrix} x+y+z\\ x+z\\ y \end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$ This gives $$y=0,\quad x+z=0.$$ So one nonzero solution is $$X=\begin{pmatrix}1\\0\\-1\end{pmatrix}.$$ --- 4. **Check option B: Does there exist a nonzero vector such that $MX=-X$?** This means $$(M+I)X=0.$$ So we check whether $M+I$ is singular. $$M+I=\begin{pmatrix} 2&1&1\\ 1&1&1\\ 0&1&1 \end{pmatrix}.$$ Compute its determinant: $$\det(M+I)=\begin{vmatrix} 2&1&1\\ 1&1&1\\ 0&1&1 \end{vmatrix}.$$ Expand along the first row: $$\det(M+I)=2\begin{vmatrix}1&1\\1&1\end{vmatrix}-1\begin{vmatrix}1&1\\0&1\end{vmatrix}+1\begin{vmatrix}1&1\\0&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}1&1\\1&1\end{vmatrix}=0, \qquad \begin{vmatrix}1&1\\0&1\end{vmatrix}=1.$$ Thus, $$\det(M+I)=2(0)-1(1)+1(1)=0.$$ Hence $M+I$ is singular, so there exists a nonzero vector $X$ such that $$(M+I)X=0 \iff MX=-X.$$ Therefore, **B is true**. We can also solve explicitly: $$MX=-X$$ with $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$ gives $$\begin{pmatrix} x+y+z\\x+z\\y \end{pmatrix}=\begin{pmatrix}-x\\-y\\-z\end{pmatrix}.$$ So, \begin{align*} x+y+z&=-x \Rightarrow 2x+y+z=0,\\ x+z&=-y \Rightarrow x+y+z=0,\\ y&=-z \Rightarrow y+z=0. \end{align*} From $y=-z$, and $x+y+z=0$, we get $x=0$. Then $2x+y+z=0$ is automatically satisfied. So one nonzero solution is $$X=\begin{pmatrix}0\\1\\-1\end{pmatrix}.$$ --- 5. **Check option D: Is $(M-2I)$ invertible?** $$M-2I=\begin{pmatrix} -1&1&1\\ 1&-2&1\\ 0&1&-2 \end{pmatrix}.$$ Compute its determinant: $$\det(M-2I)=\begin{vmatrix} -1&1&1\\ 1&-2&1\\ 0&1&-2 \end{vmatrix}.$$ Expand along the first row: \begin{align*} \det(M-2I) &=(-1)\begin{vmatrix}-2&1\\1&-2\end{vmatrix}-1\begin{vmatrix}1&1\\0&-2\end{vmatrix}+1\begin{vmatrix}1&-2\\0&1\end{vmatrix}\\ &=(-1)(4-1)-1(-2)+1(1)\\ &=-3+2+1=0. \end{align*} So $(M-2I)$ is **not invertible**. Hence, **D is false**. --- 6. **Final conclusion** The true statements are: $$\boxed{B,\ C}$$ This matches the stored correct answer.

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