CUET UG 2025 — Mathematics Algebra

Given $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$A=[10​0−1​] and $B = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$B=[02​20​]

To find $AB$AB, multiply each row of $A$A by each column of $B$B.

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Position (1,1):

Row 1 of $A$A: [1, 0] and Column 1 of $B$B: [0, 2]

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

$$= 0$$=0

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Position (1,2):

Row 1 of $A$A: [1, 0] and Column 2 of $B$B: [2, 0]

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

$$= 2$$=2

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Position (2,1):

Row 2 of $A$A: [0, -1] and Column 1 of $B$B: [0, 2]

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

$$= -2$$=−2

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Position (2,2):

Row 2 of $A$A: [0, -1] and Column 2 of $B$B: [2, 0]

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

$$= 0$$=0

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