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The total number of 3 \times 3 matrices A having entries from the set \left \{ 0,1,2,3 \right \} such that the sum of all the diagonal entries of AA^{T} is 9 , is equal to ___________.
Option: 1 0
Option: 2 8
Option: 3 88
Option: 4 766

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\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]\\ &\text {diagonal elements of }\\ &AA ^{ T }, \quad a ^{2}+ b ^{2}+ c ^{2}, d ^{2}+ e ^{2}+ f ^{2}, g ^{2}+ b ^{2}+ c ^{2}\\ &\text {Sum }=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}+i^{2}=9\\ &\text { a, } b , c , d\\ &i \in\{0,1,2,3\} \end{aligned}

\begin{array}{|l|l|l|} \hline & \text { Case } & \text { No. of Matrices } \\ \hline \text { (1) } & \text { All - 1s } & \frac{9 !}{9 !}=1 \\ \hline \text { (2) } & \begin{array}{l} \text { One } \rightarrow 3 \\ \text { remaining-0 } \end{array} & \frac{9 !}{1 ! \times 8 !}=9 \\ \hline \text { (3) } & \begin{array}{l} \text { One-2 } \\ \text { five-1s } \\ \text { three-0s } \end{array} & \frac{9 !}{1 ! \times 5 ! \times 3 !}=8 \times 63 \\ \hline \text { (4) } & \begin{array}{l} \text { two - 2's } \\ \text { one-1 } \\ \text { six-0's } \end{array} & \frac{9 !}{2 ! \times 6 !}=63 \times 4 \\ \hline \end{array}

Total no. of ways = 1 + 9 + 8 X 63 + 63 X 4 = 766.

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Suraj Bhandari

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