Question

The total number of $3\times 3$ matrices $\mathrm{A}$ having entries from the set $\left\{0,1,2,3\right\}$ such that the sum of all the diagonal entries of ${\mathrm{AA}}^{\mathrm{T}}$ is $9$, is equal to

Answer 1

Find the total number of matrices:

Let $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{d}& \mathrm{e}& \mathrm{f}\\ \mathrm{g}& \mathrm{h}& \mathrm{i}\end{array}\right]$

So ${\mathrm{A}}^{\mathrm{T}}=\left[\begin{array}{ccc}\mathrm{a}& \mathrm{d}& \mathrm{g}\\ \mathrm{b}& \mathrm{e}& \mathrm{h}\\ \mathrm{c}& \mathrm{f}& \mathrm{i}\end{array}\right]$

Multiply both the matrix,

${\mathrm{AA}}^{\mathrm{T}}=\left[\begin{array}{ccc}{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2& \mathrm{ad}+\mathrm{be}+\mathrm{cf}& \mathrm{ag}+\mathrm{bh}+\mathrm{ci}\\ \mathrm{ad}+\mathrm{be}+\mathrm{cf}& {\mathrm{d}}^2+{\mathrm{e}}^2+{\mathrm{f}}^2& \mathrm{dg}+\mathrm{eh}+\mathrm{fi}\\ \mathrm{ag}+\mathrm{bh}+\mathrm{ci}& \mathrm{dg}+\mathrm{eh}+\mathrm{fi}& {\mathrm{g}}^2+{\mathrm{h}}^2+{\mathrm{i}}^2\end{array}\right]$

Since sum of all diagonal entries is trace.

$\mathrm{Tr}\left({\mathrm{AA}}^{\mathrm{T}}\right)={\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2+{\mathrm{e}}^2+{\mathrm{f}}^2+{\mathrm{g}}^2+{\mathrm{h}}^2+{\mathrm{i}}^2$

$\Rightarrow$ $9={\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2+{\mathrm{e}}^2+{\mathrm{f}}^2+{\mathrm{g}}^2+{\mathrm{h}}^2+{\mathrm{i}}^2$

Since, $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f},\mathrm{g},\mathrm{h},\mathrm{i}\in \left\{0,1,2,3\right\}$

	Case	Number of matrices
$\left(1\right)$	All variables$=1$	$\frac{9!}{9!}=1$
$\left(2\right)$	One variable$=3$ others $=0$	$\frac{9!}{1!\times 8!}=9$
$\left(3\right)$	One variable$=2$ Five variables$=1$ Three variables$=0$	$\begin{array}{rcl}\frac{9!}{1!\times 5!\times 3!}& =& 8\times 63\\ & =& 504\end{array}$
$\left(4\right)$	One variable$=1$ Six variables$=0$ Two variables$=2$	$\begin{array}{rcl}\frac{9!}{2!\times 6!}& =& 63\times 4\\ & =& 252\end{array}$

Therefore, total number of matrices,

$$\begin{gathered} =1+9+504+252 \\ =766 \end{gathered}$$

Hence, total number of matrices possible is $766$.