Target Exam

CUET

Subject

Section B1

Chapter

Matrices

Question:

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

Options:

2

4

-4

-2

Correct Answer:

-4

Explanation:

The correct answer is Option (3) → -4 ##

We have,

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

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

$= \begin{bmatrix} -3 & 4 & 1 \end{bmatrix}$

Now, $\begin{bmatrix} -3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A$

$A = \begin{bmatrix} -3 + 0 - 1 \end{bmatrix} = \begin{bmatrix} -4 \end{bmatrix}$