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 $A^{\mathrm{T}} A$ is 6 is ____________.
Answer (integer)
5376
Solution
<p>Sum of all diagonal elements is equal to sum of square of each element of the matrix.</p>
<p>i.e., $$A = \left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{b_1}} & {{b_2}} & {{b_3}} \cr
{{c_1}} & {{c_2}} & {{c_3}} \cr
} } \right]$$</p>
<p>then ${t_r}\,(A\,.\,{A^T})$</p>
<p>$= a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 + c_1^2 + c_2^2 + c_3^2$</p>
<p>$\because$ ${a_i},{b_i},{c_i} \in \{ - 1,0,1\}$ for $i = 1,2,3$</p>
<p>$\therefore$ Exactly three of them are zero and rest are 1 or $-$1.</p>
<p>Total number of possible matrices ${}^9{C_3} \times {2^6}$</p>
<p>$= {{9 \times 8 \times 7} \over 6} \times 64$</p>
<p>$= 5376$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
This question is part of PrepWiser's free JEE Main question bank. 274 more solved questions on Matrices and Determinants are available — start with the harder ones if your accuracy is >70%.