Question

Find the number of matrices $A$ of order $3\times 2$ whose elements are from the set $\{\pm2,\pm1,0\}$, if $\mathrm{Tr}(A^TA)=5$.

• A. 310
• B. 312
• C. 320
• D. 325

Hint:

$\mathrm{Tr}(A^TA)$ equals the sum of squares of all entries of $A$. Always reduce such problems to counting valid square-sum combinations.

Answer 1

The Correct Option is B

Step 1: Interpret $\mathrm{Tr(A^TA)$.}
For any real matrix $A$, \[ \mathrm{Tr}(A^TA)=\sum (\text{squares of all entries of }A) \] Since $A$ is of order $3\times 2$, it has $6$ entries.
Step 2: Possible squares of entries.
From the set $\{\pm2,\pm1,0\}$: \[ (\pm2)^2=4,\quad (\pm1)^2=1,\quad 0^2=0 \] We need the sum of squares of the 6 entries to be $5$.
Step 3: Case-wise counting.
Case I: One entry contributes $4$ and one entry contributes $1$.
Remaining $4$ entries must be $0$. \[ 4+1=5 \] Number of ways: \[ \binom{6}{1}\binom{5}{1}\times 2 \times 2 =6\times5\times2\times2=120 \] (choices of positions and signs)
Case II: Five entries contribute $1$ each.
Remaining one entry must be $0$. \[ 1+1+1+1+1=5 \] Number of ways: \[ \binom{6}{1}\times 2^5=6\times32=192 \] Step 4: Total number of matrices.
\[ 120+192=312 \]