Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

If $A =\begin{bmatrix}0&2\\3&-4\end{bmatrix}$ and $kA =\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$, then the values of k, a, b are respectively.

Options:

-6, -12, -18

-6, 4, 9

-6, -4, -9

- 6, 12, 18

Correct Answer:

-6, -4, -9

Explanation:

We have,

$A =\begin{bmatrix}0&2\\3&-4\end{bmatrix}⇒kA =\begin{bmatrix}0&2k\\3k&-4k\end{bmatrix}$

But, $kA =\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$

$∴\begin{bmatrix}0&2k\\3k&-4k\end{bmatrix}=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$

$⇒2k = 3a, 3k = 2b, - 4k = 24 ⇒k=-6, a=-4, b = -9$