CUET UG 2025 — Mathematics Algebra

The matrix $A = [a_{ij}]_{3\times3}$A=[aij​]3×3​ is defined by:

$$a_{ij} = \begin{cases} 2i + 3j, & i < j \\ 5, & i = j \\ 3i - 2j, & i > j \end{cases}$$aij​=⎩⎨⎧​2i+3j,5,3i−2j,​i<ji=ji>j​

Computing each element of the matrix:

For $i = 1$i=1:

$a_{11} = 5$a11​=5 (since $i = j$i=j)

$a_{12} = 2(1) + 3(2) = 8$a12​=2(1)+3(2)=8 (since $i < j$i<j)

$a_{13} = 2(1) + 3(3) = 11$a13​=2(1)+3(3)=11 (since $i < j$i<j)

---

For $i = 2$i=2:

$a_{21} = 3(2) - 2(1) = 4$a21​=3(2)−2(1)=4 (since $i > j$i>j)

$a_{22} = 5$a22​=5 (since $i = j$i=j)

$a_{23} = 2(2) + 3(3) = 13$a23​=2(2)+3(3)=13 (since $i < j$i<j)

---

For $i = 3$i=3:

$a_{31} = 3(3) - 2(1) = 7$a31​=3(3)−2(1)=7 (since $i > j$i>j)

$a_{32} = 3(3) - 2(2) = 5$a32​=3(3)−2(2)=5 (since $i > j$i>j)

$a_{33} = 5$a33​=5 (since $i = j$i=j)

---

The matrix is:

$$A = \begin{bmatrix} 5 & 8 & 11 \\ 4 & 5 & 13 \\ 7 & 5 & 5 \end{bmatrix}$$A=​547​855​11135​​

---

Elements greater than 7: $8, 11, 13$8,11,13

Therefore, the number of elements greater than 7 is $3$3.