Matrices — JEE Main · 2021 · 16 Mar 2021 · Shift 1 Question with Solution

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

A A^{⊤}

is 9, is equal to

Answer: 766

Solution

Let

A = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33}

. Then

A A^{T}

has diagonal entries

(A A^{T})_{ii} = \sum_{j = 1}^{3} a_{ij}^{2}

. The sum of diagonal entries is

tr (A A^{T}) = \sum_{i = 1}^{3} \sum_{j = 1}^{3} a_{ij}^{2} = 9

. Each

a_{ij} \in {0, 1, 2, 3}

, so

a_{ij}^{2} \in {0, 1, 4, 9}

. We need to count 9-tuples

(a_{11}^{2}, \dots, a_{33}^{2})

summing to 9. This is equivalent to finding the number of ways to write

9 = x_{1} + x_{2} + \dots + x_{9}

where each

x_{i} \in {0, 1, 4, 9}

. The possible combinations are: (1) one 9, eight 0's:

(1 9) = 9

ways; (2) two 4's, one 1, six 0's:

(2 9) (1 7) = 252

ways; (3) one 4, five 1's, three 0's:

(1 9) (5 8) = 504

ways; (4) nine 1's:

1

way. Total:

9 + 252 + 504 + 1 = 766

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer