Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Matrices

Question:

If $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $B =\begin{bmatrix}0&2\\2&0\end{bmatrix}$, then the matrix AB is equal to

Options:

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

$\begin{bmatrix}0&0\\0&0\end{bmatrix}$

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

$\begin{bmatrix}2&-2\\-2&2\end{bmatrix}$

Correct Answer:

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

Explanation:

The correct answer is Option (3) → $\begin{bmatrix}0&2\\-2&0\end{bmatrix}$

Given: $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$

Compute $AB$: $AB = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} (1)(0) + (0)(2) & (1)(2) + (0)(0) \\ (0)(0) + (-1)(2) & (0)(2) + (-1)(0) \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Therefore, $AB = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$