Question

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

• A. M is invertible
• B. There exists a nonzero column matrix\(\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\) such that M\(\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\)=\(\begin{pmatrix}-a_1\\-a_2\\-a_3\end{pmatrix}\)
• C. The set  {\({x\in R^3:MX=0}\)} \(\neq\) 0, where 0=\(\begin{pmatrix}0\\0\\0\end{pmatrix}\)
• D. The matrix ( M-2I) is invertible, where I is the 3×3 identity matrix

Answer 1

The Correct Option is B, C

Given :
M = (a_ij), i, j ∈ {1, 2, 3},
a_ij = 1 if j + 1 is divisible by i, otherwise a_ij = 0
\(M=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\)
|M| = 1(-1) - 1(-1)
= -1 + 1 = 0
So, M is not invertible
\(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \end{bmatrix}=\begin{bmatrix} -a_{1} \\ -a_{2} \\ -a_{3} \end{bmatrix}\)
\(\begin{bmatrix} a_1+ a_2 +a_3 \\ a_1+a_3   \\ a_2   \end{bmatrix}=\begin{bmatrix} -a_{1} \\ -a_{2} \\ -a_{3} \end{bmatrix}\)

There exists an infinite number of possible column matrices.
\(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
    x + y + z = 0
⇒     x + z = 0
           y = 0
So, this is possible only.
\(|M-2I|=\begin{bmatrix} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{bmatrix}\)
\(=-1(3)-1(-2-1)=-3+3=0\)
So, the correct options are (B) and (C).