Q136 / 200 · AP EAMCET 2001 · Mathematics

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

  • A
    $-6,-12,-18$
  • B
    $-6,4,9$
  • Correct answer
    C
    $-6,-4,-9$
  • D
    $-6,+12,18$

Solution

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, $\begin{aligned} & 3 a=2 h \\ & \Rightarrow \quad a=\frac{2 h}{3} \\ & 3 h=2 b \\ & \Rightarrow \quad b=\frac{3 h}{2} \\ & \text { and } \quad-4 h=24 \Rightarrow h=-6 \\ & \therefore \quad a=-4, b=-9 \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.

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