Target Exam

CUET

Subject

Section B1

Chapter

Matrices

Question:

Find the values of $a$ and $b$, if $A = B$, where $A = \begin{bmatrix} a + 4 & 3b \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2a + 2 & b^2 + 2 \\ 8 & b^2 - 5b \end{bmatrix}$.

Options:

$a=2,b=2$

$a=2,b=−1$

$a=−2,b=2$

$a=−2,b=−1$

Correct Answer:

$a=2,b=2$

Explanation:

The correct answer is Option (1) → $a=2,b=2$ ##

We have, $A = \begin{bmatrix} a + 4 & 3b \\ 8 & -6 \end{bmatrix}_{2 \times 2}$ and $B = \begin{bmatrix} 2a + 2 & b^2 + 2 \\ 8 & b^2 - 5b \end{bmatrix}_{2 \times 2}$

Also, $A = B$.

For equality of matrices we know that each element of $A$ is equal to the corresponding element of $B$, that is $a_{ij} = b_{ij}$ for all $i$ and $j$.

$∴a_{11} = b_{11} \Rightarrow a + 4 = 2a + 2 \Rightarrow a = 2$

$a_{12} = b_{12} \Rightarrow 3b = b^2 + 2 \Rightarrow b^2 = 3b - 2$

and $a_{22} = b_{22} \Rightarrow -6 = b^2 - 5b$

$\Rightarrow -6 = 3b - 2 - 5b \quad [∵b^2 = 3b - 2]$

$\Rightarrow 2b = 4 \Rightarrow b = 2$

$∴ a = 2$ and $b = 2$.