JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

The number of $3 \times 2$ matrices $A$ , which can be formed using the elements of the set ${- 2, - 1, 0, 1, 2}$ such that the sum of all the diagonal elements of $A^{T} A$ is 5, is _____ .

Your answer

Solution

Let $A = a_{1} b_{1} a_{2} b_{2} a_{3} b_{3}$ . The diagonal of $A^{T} A$ has entries $a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ and $b_{1}^{2} + b_{2}^{2} + b_{3}^{2}$ , so

$tr (A^{T} A) = a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + b_{1}^{2} + b_{2}^{2} + b_{3}^{2} = 5$ .

Each entry is in ${- 2, - 1, 0, 1, 2}$ , so each square is $0$ , $1$ , or $4$ . We need six such squares summing to $5$ .

Case A: one $4$ , one $1$ , four $0$ 's.

Choose position for $\pm 2$ : 6 ways, with sign: 2 ways. Choose one of remaining 5 for $\pm 1$ : 5 ways, with sign: 2 ways.

Count: $6 \cdot 2 \cdot 5 \cdot 2 = 120$ .

Case B: five $1$ 's, one $0$ .

Choose the zero position: 6 ways. Each of the other 5 entries is $\pm 1$ : $2^{5} = 32$ .

Count: $6 \cdot 32 = 192$ .

Total: $120 + 192 = 312$ .

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