Let R = \ ( cases a & 3 & b c & 2 & d 0 & 5 & 0 cases ): a, b, c, d...

Given,

R = \left[\begin{cases}a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0\end{cases}\right]

⇒ | R | = - 5\left|\begin{cases}a & b \\ c & d\end{cases}\right|

We know that, for invertible matrices \left|R\right| \ne 0$</p><p>Now$| R |$can be zero in following cases:</p><p>(i) Two of$a, b, c, d$are zeroes which can be ($a$and$b$),($b$and$d$),($d$and$c$) or ($c$and$a$)</p><p>$\to 4 \times {7}^{2}\text{ways} = 196$</p><p>(ii) Any three of$a, b, c, d$are zeroes</p><p>$\to C3 4 \times 7 = 28$</p><p>(iii) All four of$a, b, c, d$are zeroes</p><p>$\to 1$</p><p>(iv) All four of$a, b, c, d$are non-zero but same number</p><p>$\to 7$</p><p>(v) When two are alike and$2$other are alike (non-zero)</p><p>$\to 7{C}_{2} \times 2 \times 2 = 84$</p><p>Number of invertible matrices</p><p>$= {8}^{4} - 196 - 28 - 1 - 7 - 84 = 3780