Matrices – Construction and Trace – JEE Main 24 Jan 2026 Shift 1

Solution:

Let $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix}$.
$A^TA = \begin{bmatrix} a_1^2+a_2^2+a_3^2 & \dots \\ \dots & b_1^2+b_2^2+b_3^2 \end{bmatrix}$.
Trace (sum of diagonal elements) of $A^TA = \sum a_i^2 + \sum b_i^2$.
Let $S = \sum_{i=1}^3 (a_i^2 + b_i^2) = 5$.
The elements are from $\{-2, -1, 0, 1, 2\}$. Squares are $\{4, 1, 0, 1, 4\} = \{0, 1, 4\}$.
We need to partition 5 into a sum of 6 squares (since matrix is $3 \times 2$, 6 elements).
Possible combinations of squares $\{0, 1, 4\}$ to sum to 5:

Case 1: One ‘4’, One ‘1’, Four ‘0’s.
Elements used:
– For square 4: $\pm 2$ (2 choices)
– For square 1: $\pm 1$ (2 choices)
– For square 0: $0$ (1 choice)
Total elements selection: One from $\{\pm 2\}$, One from $\{\pm 1\}$, Four $0$s.
Number of ways to arrange: $\frac{6!}{1!1!4!} \times 2 \times 2 = 30 \times 4 = 120$.

Case 2: Five ‘1’s, One ‘0’.
Elements used:
– For square 1: $\pm 1$ (2 choices for each position)
– For square 0: $0$
Total elements selection: Five from $\{\pm 1\}$, One $0$.
Number of ways to arrange positions: $\frac{6!}{5!1!} = 6$.
For each of the 5 positions with ‘1’, we have 2 choices ($\pm 1$).
Total ways: $6 \times 2^5 = 6 \times 32 = 192$.

Case 3: Two ‘4’s … sum > 5. Impossible.
Case 4: Three ‘1’s, One ‘2’ (square 4)? Sum $1+1+1+4 = 7 > 5$.

Are there other combinations?
Sum = 5.
4+1+0+0+0+0 (Handled)
1+1+1+1+1+0 (Handled)
2+1+1+1? Square of 2 is 4. 4+1+1+1 = 7. No.
3+… No 3 sq.

So total ways = $120 + 192 = 312$.

Ans. (312)