Verified Solution • 2025–26 Session moderate

QJEEAdv23Ma9704: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma9704 Solution: Let , be the matrix such that if is divisible by , otherwise .…

Let M=(aij),i,j∈{1,2,3}M=\left(a_{i j}\right), i, j \in\{1,2,3\} , be the 3×33 \times 3 matrix such that aij=1a_{i j}=1 if j+1j+1 is divisible by ii , otherwise aij=0a_{i j}=0 . Then which of the following statements is(are) true?

Multiple Choice Options

  • A MM is invertible
  • B There exists a nonzero column matrix (a1a2a3)\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right) such that M(a1a2a3)=(−a1−a2−a3)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 {X∈R3:MX=0}≠{0}\left\{X \in \mathbb{R}^3: M X=\mathbf{0}\right\} \neq\{\mathbf{0}\} , where 0=(000)\mathbf{0}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)
  • D The matrix (M−2I)(M-2 I) is invertible, where II is the 3×33 \times 3 identity matrix

Step-by-Step Verified Solution

M=[a11amp;a12amp;a13a21amp;a22amp;a23a31amp;a32amp;a33]=[1amp;1amp;11amp;0amp;10amp;1amp;0]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⇒M|M|=-1+1=0 \Rightarrow M is singular so non-invertible ⇒\Rightarrow [A] is wrong.

(B) M[a1a2a3]=[−a1−a2−a3]⇒[1amp;1amp;11amp;0amp;10amp;1amp;0][a1a2a3]=[−a1−a2−a3]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]

a1+a2+a3=−a1a1+a3=−a2a2=−a3}⇒a1=0 and a2+a3=0 infinite solutions exists [B] is correct. \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) :

amp;M−2I=[1amp;1amp;11amp;0amp;10amp;1amp;0]−2[1amp;0amp;00amp;1amp;00amp;0amp;1]=[−1amp;1amp;11amp;−2amp;10amp;1amp;−2]amp;∣M−2I∣=0⇒[D] is wrong \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) :

amp;MX=0⇒[1amp;1amp;11amp;0amp;10amp;1amp;0][xyz]=[000]amp;x+y+z=0amp;x+z=0amp;y=0amp;∴ Infinite solution \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}

[C][\mathrm{C}] is correct

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