# The number of all $3\times 3$ matrices A, with enteries from the set $\left \{ -1,0,1 \right \}$ such that the sum of the diagonal elements of $AA^{T}$ is 3, is Option: 1 672 Option: 2 512 Option: 31024 Option: 4 256

Let matrix A be

$\\A=\begin{bmatrix} a &b &c \\ d& e &f \\ g &h & i \end{bmatrix}\\\\\\ A^T=\begin{bmatrix} a &d &g \\ b& e &h \\ c &f & i \end{bmatrix} \\\\\\ \text{trace}(AA^T)=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=3$

So out of 9 elements, 3 elements must be equal to 1 or −1, and the rest elements must be 0.

Possible cases

$\begin{array}{cc} 0,0,0,0,0,0,1,1,1 & \rightarrow \text{Total possibilities}=^9C_6 \\ 0,0,0,0,0,0,-1,-1,-1 & \rightarrow \text{Total possibilities}=^9C_6 \\ 0,0,0,0,0,0,1,1,-1 & \rightarrow \text{Total possibilities}=^9C_6\times 3 \\ 0,0,0,0,0,0,-1,1,-1 & \rightarrow \text{Total possibilities}=^9C_6\times3 \end{array}$

$\\\\\text{Total number of cases}=^{9}{C}_{6}\times 8=672$

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