Question

$A$ is a $3\times 3$ diagonal matrix having integral entries such that $\text{det}\left(A\right)=120$, number of such matrices is $10n$, then $n$ is

• A. $36$
• B. $38$
• C. $34$
• D. $30$

Answer 1

The correct option is A $36$

$A=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$

$|A|=\underset{a,b,c\to +ve}{abc}=\underset{any,two\to -ve\&3rdis+ve}{120}$

$\Rightarrow (a,b,c)$

$1,1,120\Rightarrow 3!|2!2,2,3\Rightarrow 3!|2!3,4,10\Rightarrow 3!$

$1,2,60\Rightarrow 3!2,3,20\Rightarrow 3!|3,5,8\Rightarrow 3!$

$1,3,40\Rightarrow 3!2,4,15\Rightarrow 3!|3,8,5\Rightarrow 3!$

$1,4,30\Rightarrow 3!2,6,10\Rightarrow 3!4,5,6\Rightarrow 3!$

$1,6,20\Rightarrow 3!$

$1,8,15\Rightarrow 3!$

$1,10,12\Rightarrow 3!$

Case 1:(a,b,c are positive) $\Rightarrow 14(3!)+2(3!|2!)$

$=14\times 6+2\times 3$

$=90$

Case 2: (any two are -ve): ${}^3{C}_2{}^1{C}_1\left(90\right)$

$=3\times 1\times 90$

$=270$

Total matrices $=90+270=10n$

$=360=10n$

$=n=36$