JEE Mathematics 2026 Question with Solution

Given: A $3\times2$ matrix $A$ has entries from the set $\{-2,-1,0,1,2\}$, and the sum of all diagonal elements of $A^TA$ is $5$.

Find: The number of such matrices $A$.

For any matrix $A$, the sum of diagonal elements of $A^TA$ equals the trace of $A^TA$, which is the sum of squares of all entries of $A$:

So the condition becomes

The possible values of entries of $A$ are $-2,-1,0,1,2$, so their squares can only be

Since $A$ is a $3\times2$ matrix, it has $6$ entries. We need six square-values chosen from $\{4,1,0\}$ whose sum is $5$.

The only possible decomposition is

Therefore, exactly one entry must have absolute value $2$, exactly one entry must have absolute value $1$, and the remaining four entries must be $0$.

Now count the matrices.

Choose the position of the entry with absolute value $2$ in $6$ ways.

Choose the position of the entry with absolute value $1$ from the remaining $5$ places in $5$ ways.

So the number of position choices is

Since the entry with absolute value $2$ can be $2$ or $-2$, and the entry with absolute value $1$ can be $1$ or $-1$, the sign choices are

Hence the total number of matrices is

Therefore, the required number of matrices is $36$.

Given: The trace of $A^TA$ is $5$.

Find: The number of $3\times2$ matrices satisfying this condition.

Use the fact that $\text{trace}(A^TA)$ is the sum of squares of all six entries of $A$. Since the allowed squares are only $0,1,4$ and the total is $5$, the pattern is forced to be one $4$, one $1$, and four $0$'s.

Choose the two non-zero positions:

Among these two chosen positions, one must contain the entry with absolute value $2$ and the other with absolute value $1$. Their magnitudes are distinguished by the sum condition, so this assignment is built into the positional count when counted as ordered placement or equivalently handled directly as one position for $|2|$ and one for $|1|$.

Then choose signs independently:

Thus,

Therefore, the required answer is $36$.