JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 21)
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
766
Explanation
$$A{A^T} = \left[ {\matrix{
x & y & z \cr
a & b & c \cr
d & e & f \cr
} } \right]\left[ {\matrix{
x & a & d \cr
y & b & e \cr
z & c & f \cr
} } \right]$$
$$ = \left[ {\matrix{ {{x^2} + {y^2} + {z^2}} & {ax + by + cz} & {dx + ey + fz} \cr {ax + by + cz} & {{a^2} + {b^2} + {c^2}} & {ad + be + cf} \cr {dx + ey + fz} & {ad + be + cf} & {{d^2} + {e^2} + {f^2}} \cr } } \right]$$
$$Tr(A{A^T}) = {x^2} + {y^2} + {z^2} + {a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} = 9$$
Case-I : Nine ones = 1 case
Case-II : 8 zeroes and one entry is 3 = $${{{9!} \over {8!}} = 9}$$ cases
Case-III : Two 2’s, one 1’s and 6 zeroes = $${{9!} \over {2!6!}} = 63 \times 4 = 252$$
Case IV : one 2, five 1, rest 0 $${{9!} \over {5!3!}} = 63 \times 8 = 504$$
$$ \therefore $$ Total cases = 9 + 252 + 504 + 1 = 766
$$ = \left[ {\matrix{ {{x^2} + {y^2} + {z^2}} & {ax + by + cz} & {dx + ey + fz} \cr {ax + by + cz} & {{a^2} + {b^2} + {c^2}} & {ad + be + cf} \cr {dx + ey + fz} & {ad + be + cf} & {{d^2} + {e^2} + {f^2}} \cr } } \right]$$
$$Tr(A{A^T}) = {x^2} + {y^2} + {z^2} + {a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} = 9$$
Case-I : Nine ones = 1 case
Case-II : 8 zeroes and one entry is 3 = $${{{9!} \over {8!}} = 9}$$ cases
Case-III : Two 2’s, one 1’s and 6 zeroes = $${{9!} \over {2!6!}} = 63 \times 4 = 252$$
Case IV : one 2, five 1, rest 0 $${{9!} \over {5!3!}} = 63 \times 8 = 504$$
$$ \therefore $$ Total cases = 9 + 252 + 504 + 1 = 766
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