Let the matrix \(A = [a_{ij}]_{3\times 3}\) be defined by
\(a_{ij}=\begin{cases}2i+3j, & i \lt j\\ 5, & i=j\\ 3i-2j, & i \gt j\end{cases}\)
The number of elements in the matrix A which are greater than 7, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
Above the diagonal (i < j), \(a_{ij} = 2i + 3j\):
\(a_{12} = 2 + 6 = 8\)
\(a_{13} = 2 + 9 = 11\)
\(a_{23} = 4 + 9 = 13\)
On the diagonal, \(a_{11} = a_{22} = a_{33} = 5\).
Below the diagonal (i > j), \(a_{ij} = 3i - 2j\):
\(a_{21} = 6 - 2 = 4\)
\(a_{31} = 9 - 2 = 7\)
\(a_{32} = 9 - 4 = 5\)
\( \displaystyle A = \begin{bmatrix}5&8&11\\4&5&13\\7&5&5\end{bmatrix} \)
The elements greater than 7 are 8, 11 and 13; \(a_{31} = 7\) is not greater than 7. So there are 3 such elements.