The number of all 3 × 3 matrices A, with enteries from the set –1, 0, 1 such that the sum of the diagonal elements of AAT is 3, is

Matrices and Determinants PYQ — JEE Main 2020 · 8 Jan · Shift 1 · Q23 with Solution

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Correct answer: 672

Let the rows of the $3\times 3$ matrix $A$ be $r_1=(a_{11},a_{12},a_{13}),\quad r_2=(a_{21},a_{22},a_{23}),\quad r_3=(a_{31},a_{32},a_{33}),$ where each entry belongs to $\{-1,0,1\}$ .
We need the sum of diagonal elements of $AA^T$ .

Recall that for any matrix $A$ ,
$r_1\cdot r_1 & r_1\cdot r_2 & r_1\cdot r_3\\ r_2\cdot r_1 & r_2\cdot r_2 & r_2\cdot r_3\\ r_3\cdot r_1 & r_3\cdot r_2 & r_3\cdot r_3 \end{pmatrix}.$$ Hence, $$\operatorname{tr}(AA^T)=r_1\cdot r_1+r_2\cdot r_2+r_3\cdot r_3.$$$
Since each entry is from $\{-1,0,1\}$ , we have $a_{ij}^2 \in \{0,1\}.$ Therefore, $\operatorname{tr}(AA^T)=\sum_{i,j} a_{ij}^2,$ which is simply the total number of nonzero entries in $A$ .
The condition says $\operatorname{tr}(AA^T)=3.$ So the matrix must have exactly $3$ nonzero entries.
Now count such matrices:
- Total number of positions in a $3\times 3$ matrix = $9$ .
- Choose exactly $3$ positions to be nonzero: $\binom{9}{3}=84.$
- Each chosen nonzero position can be either $1$ or $-1$ : $2^3=8.$
Hence total number of matrices is $\binom{9}{3}\cdot 2^3=84\cdot 8=672.$
Therefore, the required number is $\boxed{672}.$

View written solutionFree

Correct answer: 672

Let the rows of the $3\times 3$ matrix $A$ be $r_1=(a_{11},a_{12},a_{13}),\quad r_2=(a_{21},a_{22},a_{23}),\quad r_3=(a_{31},a_{32},a_{33}),$ where each entry belongs to $\{-1,0,1\}$ .
We need the sum of diagonal elements of $AA^T$ .

Recall that for any matrix $A$ ,
$r_1\cdot r_1 & r_1\cdot r_2 & r_1\cdot r_3\\ r_2\cdot r_1 & r_2\cdot r_2 & r_2\cdot r_3\\ r_3\cdot r_1 & r_3\cdot r_2 & r_3\cdot r_3 \end{pmatrix}.$$ Hence, $$\operatorname{tr}(AA^T)=r_1\cdot r_1+r_2\cdot r_2+r_3\cdot r_3.$$$
Since each entry is from $\{-1,0,1\}$ , we have $a_{ij}^2 \in \{0,1\}.$ Therefore, $\operatorname{tr}(AA^T)=\sum_{i,j} a_{ij}^2,$ which is simply the total number of nonzero entries in $A$ .
The condition says $\operatorname{tr}(AA^T)=3.$ So the matrix must have exactly $3$ nonzero entries.
Now count such matrices:
- Total number of positions in a $3\times 3$ matrix = $9$ .
- Choose exactly $3$ positions to be nonzero: $\binom{9}{3}=84.$
- Each chosen nonzero position can be either $1$ or $-1$ : $2^3=8.$
Hence total number of matrices is $\binom{9}{3}\cdot 2^3=84\cdot 8=672.$
Therefore, the required number is $\boxed{672}.$