Question
If $\left[\begin{array}{cc}3 c+6 & a-d \\ a+d & 2-3 b\end{array}\right]=\left[\begin{array}{cc}12 & 2 \\ -8 & -4\end{array}\right]$ then find $a b-c d$ :

Answer

(A)
$3 c+6=12 \Rightarrow c=2$....(1)
$a-d=2$....(2)
$a+d=-8$....(3)
$
2-3 b=-4 ......(4)
$
On adding equation (2) and (3), $2 a=6 \Rightarrow a=-3$
Put the value is equation (2)
$-3-d=2 \Rightarrow d=-5
$
From equation (4)
$3 b=6$
$\Rightarrow \quad b=2$
then $a b-c d=(-3)(2)-(2)(-5)=-6+10=4$

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