A) 198
B) 162
C) 126
D) 135
Correct Answer: A
Solution :
| Let the matrix \[M=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\] |
| \[\therefore \,{{M}^{T}}=\left[ \begin{matrix} {{a}_{^{11}}} & {{a}_{^{21}}} & {{a}_{^{31}}} \\ {{a}_{^{12}}} & {{a}_{^{22}}} & {{a}^{_{32}}} \\ {{a}_{^{13}}} & {{a}_{^{32}}} & {{a}_{^{33}}} \\ \end{matrix} \right]\] |
| Sum of diagonal entries of \[{{M}^{T}}M=\sum a_{ij}^{2}=5\] Possible cases are 1, 2, 0, 0, 0, 0, 0, 0, 0 which gives \[\frac{9!}{7!}\] |
| matrices = 72 and 1, 1,1, 1, 1, 0, 0, 0, 0 which gives \[\frac{9!}{5!4!}\] matrices = 126 |
| Total matrices = 72 + 126 = 198 |
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