Question

A is a $3\times 3$ diagonal having entries such that det (A) =120, number of such matrices is 10n, then n is

Answer 1

Since matrix is diagonal matrix

So, det $\left(A\right)=abc=120=2^3\times 3^1\times 5^1$

Case $I$ $\Rightarrow$ All are the $\Rightarrow 5_{C_2}\times 3_{C_2}\times 3_{C_2}=90$

Case $II$ $\Rightarrow$ One positive and two negative

$\Rightarrow 3\times (5_{C_2}\times 3_{C_2}\times 3_{C_2})=270$

Total $=90+270=360=36\times 10=36n$

So, $n=36$