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Let the matrix A=[aij]3×3A = [a_{ij}]_{3\times3} be defined by

aij={2i+3j,i<j5,i=j3i−2j,i>ja_{ij} = \begin{cases} 2i + 3j, & i < j \\ 5, & i = j \\ 3i - 2j, & i > j \end{cases}

The number of elements in the matrix A which are greater than 7, is:

Solution

✅ Correct Option: 2

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

aij={2i+3j,i<j5,i=j3i−2j,i>ja_{ij} = \begin{cases} 2i + 3j, & i < j \\ 5, & i = j \\ 3i - 2j, & i > j \end{cases}

Computing each element of the matrix:

For i=1i = 1:

a11=5a_{11} = 5 (since i=ji = j)

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

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


For i=2i = 2:

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

a22=5a_{22} = 5 (since i=ji = j)

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


For i=3i = 3:

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

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

a33=5a_{33} = 5 (since i=ji = j)


The matrix is:

A=[58114513755]A = \begin{bmatrix} 5 & 8 & 11 \\ 4 & 5 & 13 \\ 7 & 5 & 5 \end{bmatrix}


Elements greater than 7: 8,11,138, 11, 13

Therefore, the number of elements greater than 7 is 33.

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