Let $A = [a_{ij}]_{3 \times 3}$ be a matrix, defined by $a_{ij} = \begin{cases} 2i+3j & , i < j \\6 &, i=j\\ 3i-2j & , i > j \end{cases}$. The number of elements in A which are greater than 6, is

The matrix has 9 positions (3 rows × 3 columns). For each position $(i,j)$:

• Diagonal elements $(i = j)$: $a_{ij} = 6$
• Above diagonal $(i < j)$: $a_{ij} = 2i + 3j$
• Below diagonal $(i > j)$: $a_{ij} = 3i - 2j$

---

For Row 1 $(i = 1)$:

$a_{11}$: $i = j$, so $a_{11} = 6$

$a_{12}$: $i < j$ $(1 < 2)$, so $a_{12} = 2(1) + 3(2) = 2 + 6 = 8$

$a_{13}$: $i < j$ $(1 < 3)$, so $a_{13} = 2(1) + 3(3) = 2 + 9 = 11$

---

For Row 2 $(i = 2)$:

$a_{21}$: $i > j$ $(2 > 1)$, so $a_{21} = 3(2) - 2(1) = 6 - 2 = 4$

$a_{22}$: $i = j$, so $a_{22} = 6$

$a_{23}$: $i < j$ $(2 < 3)$, so $a_{23} = 2(2) + 3(3) = 4 + 9 = 13$

---

For Row 3 $(i = 3)$:

$a_{31}$: $i > j$ $(3 > 1)$, so $a_{31} = 3(3) - 2(1) = 9 - 2 = 7$

$a_{32}$: $i > j$ $(3 > 2)$, so $a_{32} = 3(3) - 2(2) = 9 - 4 = 5$

$a_{33}$: $i = j$, so $a_{33} = 6$

---

The complete matrix is:

$A = \begin{bmatrix} 6 & 8 & 11 \\ 4 & 6 & 13 \\ 7 & 5 & 6 \end{bmatrix}$

---

Elements greater than 6:

$a_{12} = 8$ ✓

$a_{13} = 11$ ✓

$a_{23} = 13$ ✓

$a_{31} = 7$ ✓

The number of elements greater than 6 is $4$.