If $A = [a_{ij}]_{2×2}$ where $a_{ij} =\left\{\begin{matrix}1,&i≠j\\0,&i=j\end{matrix}\right.$, and $I$ is the identity matrix of order 2, then $(A^2-3A +4I)$ is

(A) Symmetric Matrix
(B) Skew-symmetric Matrix
(C) Non-singular Matrix
(D) Square Matrix

Choose the correct answer from the options given below:

(A), (C) and (D) only

The correct answer is Option (3) → (A), (C) and (D) only

Given

$A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ since $a_{ij}=1$ for $i\ne j$ and $a_{ii}=0$

Compute $A^2$

$A^2=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} =\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$

Now

$A^2-3A+4I=I-3A+4I$

$=5I-3A$

$=5\begin{pmatrix}1&0\\0&1\end{pmatrix}-3\begin{pmatrix}0&1\\1&0\end{pmatrix}$

$=\begin{pmatrix}5&0\\0&5\end{pmatrix}-\begin{pmatrix}0&3\\3&0\end{pmatrix} =\begin{pmatrix}5&-3\\-3&5\end{pmatrix}$

Transpose

$\begin{pmatrix}5&-3\\-3&5\end{pmatrix}^T=\begin{pmatrix}5&-3\\-3&5\end{pmatrix}$

Hence matrix is symmetric

Determinant

$|5I-3A|=25-9=16\ne0$

Hence matrix is non-singular

It is clearly a square matrix of order $2$

The correct options are (A), (C) and (D).