Target Exam

CUET

Subject

Section B1

Chapter

Matrices

Question:

If $\begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix} = P + Q$, where $P$ is a symmetric and $Q$ is a skew symmetric matrix, then $Q$ is equal to:

Options:

$\begin{bmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & 4 \end{bmatrix}$

$\begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$

$\begin{bmatrix} 0 & \frac{5}{2} \\ -\frac{5}{2} & 0 \end{bmatrix}$

$\begin{bmatrix} 2 & -\frac{5}{2} \\ \frac{5}{2} & 4 \end{bmatrix}$

Correct Answer:

$\begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$

Explanation:

The correct answer is Option (2) → $\begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$ ##

$\begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix} = P + Q = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

$2A = \begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix}$

$2A^T = \begin{bmatrix} 2 & 5 \\ 0 & 4 \end{bmatrix}$

$Q = \frac{A - A^T}{2}$

$= \frac{1}{2} \left( \begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 5 \\ 0 & 4 \end{bmatrix} \right)$

$= \frac{1}{2} \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$