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If $A=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right], h A=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$, then the values of $h, a, b$ are respectively
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Verified Answer
The correct answer is:
$-6,-4,-9$
We have,
$\begin{aligned} A & =\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right] \\ \Rightarrow \quad h A & =\left[\begin{array}{cc}0 & 2 h \\ 3 h & -4 h\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]\end{aligned}$
On comparing the corresponding elements,
$\begin{aligned} & 3 a=2 h \\ & \Rightarrow \quad a=\frac{2 h}{3} \\ & 3 h=2 b \\ & \Rightarrow \quad b=\frac{3 h}{2} \\ & \text { and } \quad-4 h=24 \Rightarrow h=-6 \\ & \therefore \quad a=-4, b=-9 \\ & \end{aligned}$
$\begin{aligned} A & =\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right] \\ \Rightarrow \quad h A & =\left[\begin{array}{cc}0 & 2 h \\ 3 h & -4 h\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]\end{aligned}$
On comparing the corresponding elements,
$\begin{aligned} & 3 a=2 h \\ & \Rightarrow \quad a=\frac{2 h}{3} \\ & 3 h=2 b \\ & \Rightarrow \quad b=\frac{3 h}{2} \\ & \text { and } \quad-4 h=24 \Rightarrow h=-6 \\ & \therefore \quad a=-4, b=-9 \\ & \end{aligned}$
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