Let \(A = [a_{ij}]\) be the square matrix of order 3 in which \(a_{ij} = 1\) if i divides j, and \(a_{ij} = 0\) otherwise. The sum of all the elements of \(A^2\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 7
Among 1, 2 and 3: 1 divides all three, 2 divides only 2, and 3 divides only 3. So
\( \displaystyle A = \begin{bmatrix}1&1&1\\0&1&0\\0&0&1\end{bmatrix} \)
Multiplying rows of A by columns of A:
Row 1: \(\begin{bmatrix}1&1 + 1&1 + 1\end{bmatrix} = \begin{bmatrix}1&2&2\end{bmatrix}\)
Row 2: \(\begin{bmatrix}0&1&0\end{bmatrix}\)
Row 3: \(\begin{bmatrix}0&0&1\end{bmatrix}\)
The sum of all the elements of \(A^2\) is
\(1 + 2 + 2 + 1 + 1 = 7\)
Why the other options are incorrect
Option (1) 5 — Incorrect. It is the sum of the elements of A itself, not of \(A^2\).
Option (3) 9 — Incorrect. It is the number of elements of \(A^2\), not their sum.
Option (4) 6 — Incorrect. The non-zero elements of \(A^2\) are 1, 2, 2, 1, 1, which add to 7.