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If A=[100−1]A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} and B=[0220]B = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}, then the matrix AB is equal to

Solution

✅ Correct Option: 3

Given A=[100−1]A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} and B=[0220]B = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}

To find ABAB, multiply each row of AA by each column of BB.


Position (1,1):

Row 1 of AA: [1, 0] and Column 1 of BB: [0, 2]

(1×0)+(0×2)(1 \times 0) + (0 \times 2)

=0= 0


Position (1,2):

Row 1 of AA: [1, 0] and Column 2 of BB: [2, 0]

(1×2)+(0×0)(1 \times 2) + (0 \times 0)

=2= 2


Position (2,1):

Row 2 of AA: [0, -1] and Column 1 of BB: [0, 2]

(0×0)+(−1×2)(0 \times 0) + (-1 \times 2)

=−2= -2


Position (2,2):

Row 2 of AA: [0, -1] and Column 2 of BB: [2, 0]

(0×2)+(−1×0)(0 \times 2) + (-1 \times 0)

=0= 0


Therefore, AB=[02−20]AB = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}

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