Medium INTEGER +4 / -1 PYQ · JEE Mains 2021

The total number of 3 $\times$ 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AAT is 9, is equal to _____________.

Answer (integer) 766

Solution

$$A{A^T} = \left[ {\matrix{ x &amp; y &amp; z \cr a &amp; b &amp; c \cr d &amp; e &amp; f \cr } } \right]\left[ {\matrix{ x &amp; a &amp; d \cr y &amp; b &amp; e \cr z &amp; c &amp; f \cr } } \right]$$<br><br>$$ = \left[ {\matrix{ {{x^2} + {y^2} + {z^2}} &amp; {ax + by + cz} &amp; {dx + ey + fz} \cr {ax + by + cz} &amp; {{a^2} + {b^2} + {c^2}} &amp; {ad + be + cf} \cr {dx + ey + fz} &amp; {ad + be + cf} &amp; {{d^2} + {e^2} + {f^2}} \cr } } \right]$$<br><br>$$Tr(A{A^T}) = {x^2} + {y^2} + {z^2} + {a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} = 9$$<br><br>Case-I : Nine ones = 1 case<br><br>Case-II : 8 zeroes and one entry is 3 = ${{{9!} \over {8!}} = 9}$ cases<br><br>Case-III : Two 2’s, one 1’s and 6 zeroes = ${{9!} \over {2!6!}} = 63 \times 4 = 252$<br><br>Case IV : one 2, five 1, rest 0 ${{9!} \over {5!3!}} = 63 \times 8 = 504$<br><br>$\therefore$ Total cases = 9 + 252 + 504 + 1 = 766

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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