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Question:

If $A=\left[\begin{array}{rr}0 & 2 \\ 3 & -4\end{array}\right]$ and $k A=\left[\begin{array}{rr}0 & 3 a \\ 2 b & 24\end{array}\right]$, then the values of $k, a, b$, are respectively

(a) $-6,-12,-18$

(b) $-6,4,9$

(c) $-6,-4,-9$

(d) $-6,12,18$

Solution:

(c) $-6,-4,-9$

Given : $A=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]$

Here,

$k A=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$

$\Rightarrow k\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}0 & 2 k \\ 3 k & -4 k\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\Rightarrow 2 k=3 a, 3 k=2 b$ and $-4 k=24$

Now,

$-4 k=24$

$\Rightarrow k=-6$

Also,

$2 k=3 a$ and $3 \mathrm{k}=2 b$

$\Rightarrow 2(-6)=3 a$ and $3(-6)=2 b \quad[$ Using $k=-6]$

$\Rightarrow-12=3 a$ and $-18=2 b$

$\therefore a=-4$ and $b=-9$

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