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If A=[aij]A = [a_{ij}] is a square matrix of order 2 such that aij={2,when i≠j0,when i=ja_{ij} = \begin{cases} 2, & \text{when } i \neq j \\ 0, & \text{when } i = j \end{cases}, then det (A2)(A^2) is:

Solution

✅ Correct Option: 4

For a 2×2 matrix A where each element aija_{ij} follows the given rule:

  • When i≠ji \neq j (off-diagonal): aij=2a_{ij} = 2
  • When i=ji = j (diagonal): aij=0a_{ij} = 0

A=[a11a12a21a22]A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}

For each element:

a11a_{11}: Here i=1,j=1i = 1, j = 1 → i=ji = j → a11=0a_{11} = 0

a12a_{12}: Here i=1,j=2i = 1, j = 2 → i≠ji \neq j → a12=2a_{12} = 2

a21a_{21}: Here i=2,j=1i = 2, j = 1 → i≠ji \neq j → a21=2a_{21} = 2

a22a_{22}: Here i=2,j=2i = 2, j = 2 → i=ji = j → a22=0a_{22} = 0

Therefore:

A=[0220]A = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}


For a 2×2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, determinant =ad−bc= ad - bc

det⁡(A)=(0)(0)−(2)(2)\det(A) = (0)(0) - (2)(2)

det⁡(A)=0−4\det(A) = 0 - 4

det⁡(A)=−4\det(A) = -4


Using the property det⁡(A2)=[det⁡(A)]2\det(A^2) = [\det(A)]^2:

det⁡(A2)=(−4)2\det(A^2) = (-4)^2

det⁡(A2)=16\det(A^2) = 16

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