Question

The number of $3\times2$ 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

Hint:

For $A^TA$, remember that the trace equals the sum of squares of all elements of $A$.

Answer 1

Correct Answer: 36

Step 1: Understanding the condition.
For any matrix $A$, the sum of diagonal elements of $A^TA$ is \[ \text{trace}(A^TA)=\sum \text{(squares of all elements of $A$)} \] Hence, the given condition implies \[ \sum a_{ij}^2 = 5 \] Step 2: Possible square values.
From the set $\{-2,-1,0,1,2\}$, squares are \[ \{4,1,0\} \] We need combinations of six entries (since $3\times2$ matrix) whose squares sum to $5$.
The only possible way is \[ 5=4+1 \] Thus, exactly one entry has absolute value $2$ and one entry has absolute value $1$, rest are $0$.
Step 3: Counting arrangements.
Number of ways to choose positions: \[ \binom{6}{2}=15 \] Each non-zero entry can be positive or negative: \[ 2\times2=4 \] \[ \text{Total matrices}=15\times4=36 \]