Question

The total number of 3 × 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AA^T is 9, is equal to ______.

विकल्प

• 764

• 765

• 766

• 767

Answer 1

The total number of 3 × 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AA^T is 9, is equal to 766.

Explanation:

Given that order of matrix A is 3 × 3,

Let A = `[(a, b, c),(d, e, f),(g, h, i)]_(3 xx 3)`

Then A^T = `[(a, d, g),(b, e, h),(c, f, i)]_(3 xx 3)`

AA^T = `[(a, b, c),(d, e, f),(g, h, i)][(a, d, g),(b, e, h),(c, f, i)]`

⇒ AA^T = `[(a^2 + b^2 + c^2, ad + be + cf, ag + bh + ci),(ad + be + fc, d^2 + e^2 + f^2, dg + eh + fi),(ga + hb + ci, gd + he + fi, g^2 + h^2 + i^2)]`

Now, sum of all the diagonal entries of AA^T = 9

⇒ T_r(AA^T) = 9

⇒ a² + b² + c² + d² + e² + f² + g² + h² + i² = 9  ...(i)

Given that matrix A having entries from the set {0, 1, 2, 3}

Hence a, b, c, d, e, f, g, h, i, ∈{0, 1, 2, 3}

S.N.	Case	Number of matrices
1	All→1s	`(9!)/(9!)` = 1
2	One→3, Remaining→0s	`(9!)/(1!8!)` = 9
3	One→2 Five→1s Three→0s	`(9!)/(1!5!3!)` = 504
4	Two→2s One→1 Six→0s	`(9!)/(2!6!)` = 252

So total number of ways = `(9!)/(9!) = (9!)/(1!8!) + (9!)/(1!5!3!) + (9!)/(2!6!)` = 1 + 9 + 504 + 252 = 766.