Target Exam

CUET

Subject

Section B1

Chapter

Matrices

Question:

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$, then find $2A - B$.

Options:

$\begin{bmatrix} 4 & 1 & 6 \\ 1 & 3 & 3 \end{bmatrix}$

$\begin{bmatrix} 1 & -5 & -3 \\ -5 & -6 & 0 \end{bmatrix}$

$\begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$

$\begin{bmatrix} -1 & 3 & 3 \\ 3 & 6 & 0 \end{bmatrix}$

Correct Answer:

$\begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$

Explanation:

The correct answer is Option (3) → $\begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$ ##

We have

$2A - B = 2 \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} - \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$

$= \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix} + \begin{bmatrix} -3 & 1 & -3 \\ 1 & 0 & -2 \end{bmatrix}$

$= \begin{bmatrix} 2 - 3 & 4 + 1 & 6 - 3 \\ 4 + 1 & 6 + 0 & 2 - 2 \end{bmatrix} = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$