JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 2)
The number of all 3 × 3 matrices A, with
enteries from the set {–1, 0, 1} such that the sum
of the diagonal elements of AAT is 3, is
Answer
672
Explanation
Let A = $$\left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$
$$ \therefore $$ AT = $$\left[ {\matrix{ {{a_{11}}} & {{a_{21}}} & {{a_{31}}} \cr {{a_{12}}} & {{a_{22}}} & {{a_{32}}} \cr {{a_{13}}} & {{a_{23}}} & {{a_{33}}} \cr } } \right]$$
diagonal elements of AAT are $$a_{11}^2 + a_{12}^2 + a_{13}^2$$ ,
$$a_{21}^2 + a_{22}^2 + a_{23}^2$$ , $$a_{31}^2 + a_{32}^2 + a_{33}^2$$
Given Sum = ($$a_{11}^2 + a_{12}^2 + a_{13}^2$$) +
($$a_{21}^2 + a_{22}^2 + a_{23}^2$$) + ($$a_{31}^2 + a_{32}^2 + a_{33}^2$$) = 3
This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0.
$$ \therefore $$ Number of matrices = 9C3 $$ \times $$ 2 $$ \times $$ 2 $$ \times $$ 2
= 672
$$ \therefore $$ AT = $$\left[ {\matrix{ {{a_{11}}} & {{a_{21}}} & {{a_{31}}} \cr {{a_{12}}} & {{a_{22}}} & {{a_{32}}} \cr {{a_{13}}} & {{a_{23}}} & {{a_{33}}} \cr } } \right]$$
diagonal elements of AAT are $$a_{11}^2 + a_{12}^2 + a_{13}^2$$ ,
$$a_{21}^2 + a_{22}^2 + a_{23}^2$$ , $$a_{31}^2 + a_{32}^2 + a_{33}^2$$
Given Sum = ($$a_{11}^2 + a_{12}^2 + a_{13}^2$$) +
($$a_{21}^2 + a_{22}^2 + a_{23}^2$$) + ($$a_{31}^2 + a_{32}^2 + a_{33}^2$$) = 3
This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0.
$$ \therefore $$ Number of matrices = 9C3 $$ \times $$ 2 $$ \times $$ 2 $$ \times $$ 2
= 672
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