Question

Let S be the set containing all $3\times 3$ matrices with entries from {$–1,0,1$}. The total number of matrices $A\in S$ such that the sum of all the diagonal elements of $A^{T}A$ is $6$ is ​​​​​​​

• A. 5376.0
• B. 5376
• C. 5376.00

Answer 1

Answer: (A), (B), (C)

Sum of all diagonal elements is equal to sum of square of each element of the matrix.

i.e, $A=\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right]$

then $t_r(A,A^T)$

$=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2+c_1^2+c_2^2+c_3^2$

$\therefore a_i,b_i,c_iϵ\text{{–1, 0, 1}}$ for $i=1,2,3$

$\therefore$ Exactly three of them are zero and rest are $1$ or$-1$

Total number of possible matrices ${}^9{C}_3\times 2^6$

$=\frac{9\times 8\times 7}{6}\times 64$

$=5376$



Let S be the set containing all $3\times 3$ matrices with entries from {$–1,0,1$}. The total number of matrices $A\in S$ such that the sum of all the diagonal elements of $A^{T}A$ is $6$ is _