Q114 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
Given $A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$ and $f(x) = x^2 - 2x - 3$ then $f(A)$ is:
- ANull matrix
- Correct answerBSymmetric Matrix
- CSkew symmetric matrix
- DIdentity matrix
Solution
Given $A = \begin{pmatrix} 1 & 2 \\\\ 2 & 3 \end{pmatrix}$ and $f(x) = x^2 - 2x - 3$
$f(A) = A^2 - 2A - 3I$
$A^2 = \begin{pmatrix} 1 & 2 \\\\ 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\\\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 8 \\\\ 8 & 13 \end{pmatrix}$
$f(A) = \begin{pmatrix} 5 & 8 \\\\ 8 & 13 \end{pmatrix} - \begin{pmatrix} 2 & 4 \\\\ 4 & 6 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\\\ 0 & 3 \end{pmatrix}$
$f(A) = \begin{pmatrix} 0 & 4 \\\\ 4 & 4 \end{pmatrix}$
Since $(f(A))^T = f(A)$, the resulting matrix is a symmetric matrix.
Answer: Symmetric Matrix
$f(A) = A^2 - 2A - 3I$
$A^2 = \begin{pmatrix} 1 & 2 \\\\ 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \\\\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 5 & 8 \\\\ 8 & 13 \end{pmatrix}$
$f(A) = \begin{pmatrix} 5 & 8 \\\\ 8 & 13 \end{pmatrix} - \begin{pmatrix} 2 & 4 \\\\ 4 & 6 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\\\ 0 & 3 \end{pmatrix}$
$f(A) = \begin{pmatrix} 0 & 4 \\\\ 4 & 4 \end{pmatrix}$
Since $(f(A))^T = f(A)$, the resulting matrix is a symmetric matrix.
Answer: Symmetric Matrix
Hence, the correct answer is option: 'B'.