Question

Let $\mathrm{M}$ be any $3\times 3$ matrix with entries from the set $\{0,1,2\}$ .The maximum number of such matrices, for which the sum of diagonal elements of $M^{T}M$ is seven, is

Answer 1

Finding the maximum number of matrix:

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

Given the sum of diagonal elements of $M^{T}M$ is seven,

$$\begin{gathered} \Rightarrow \left[M\right]\times \left[M^T\right]=7\left[\text{Given}\right] \\ \Rightarrow \left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\times \left[\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right]=7 \\ \Rightarrow a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=7 \end{gathered}$$

Case I: Seven $(1\text{'}s)$ and Two $(0\text{'}s)$

${\Rightarrow }^9{C}_2=36$

Case II: One $(2\text{'}s)$ and Three $(1\text{'}s)$ and five $(0\text{'}s)$

$\Rightarrow \frac{9!}{5!3!}=504$

Therefore, total $=$Case I $+$ Case II

$$\begin{gathered} =36+504 \\ =540 \end{gathered}$$

Hence, the maximum number of such matrices, for which the sum of diagonal elements of $M^{T}M$ is seven, is $540$.