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