JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Matrices

Writing $A=\left(\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\\ a_3& b_3\end{array}\right)$, the diagonal entries of $A^{T}A$ are $a_1^2+a_2^2+a_3^2$ and $b_1^2+b_2^2+b_3^2$, so their sum is the sum of the squares of all six entries. The requirement is that six numbers chosen from $\{-2,-1,0,1,2\}$ have squares adding to $5$. The available squares are $0,1,4$, so the only ways to reach $5$ are $4+1+0+0+0+0$ and $1+1+1+1+1+0$. For $4+1+0+0+0+0$, choose the position of the $\pm 2$ entry ($6$ ways, $2$ signs) and the position of the $\pm 1$ entry among the rest ($5$ ways, $2$ signs), giving $6\cdot 2\cdot 5\cdot 2=120$. For five $\pm 1$ entries and one $0$, choose the position of the $0$ ($6$ ways) and a sign for each of the five $\pm 1$ entries ($2^5=32$), giving $6\cdot 32=192$. The number of such matrices is $120+192=312$.