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Let $R=\left\{\left(\begin{array}{lll}\mathrm{a} & 3 & \mathrm{~b} \\ \mathrm{c} & 2 & \mathrm{~d} \\ 0 & 5 & 0\end{array}\right): \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in\{0,3,5,7,11,13,17,19\}\right\}$. Then the number of invertible matrices in $\mathrm{R}$ is
$500$
$3780$
$515$
$520$
Solution
Let us calculate when $|R|=0$
Case-I $a d=b c=0$
Now $\mathrm{ad}=0$
$\Rightarrow$ Total – (When none of a & $d$ is 0 )
$=8^2-1=15$ ways
Similarly bc $=0 \Rightarrow 15$ ways
$\therefore 15 \times 15=225$ ways of $a d=b c=0$
Case-II $a d=b c \neq 0$
either $a=d=b=c \quad$ OR $\quad a \neq d, b \neq d$ but $a d=b c$
${ }^7 \mathrm{C}_1=7$ ways
${ }^7 \mathrm{C}_2 \times 2 \times 2=84$ ways
Total 91 ways
$\therefore|\mathbb{R}|=0 \text { in } 225+91=316 \text { ways }$
$|\mathbb{R}| \neq 0 \text { in } 8^4-316=3780$