CUET UG 2025 — Mathematics Algebra

Given $A^T = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$AT=[−21​32​] and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$B=[−11​02​].

To find $(A + 2B)^T$(A+2B)T, matrix $A$A is needed first.

Since $A^T$AT is given, transpose it to find $A$A:

$$A = (A^T)^T$$A=(AT)T

$$A = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}^T$$A=[−21​32​]T

$$A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$$A=[−23​12​]

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Calculate $2B$2B:

$$2B = 2 \times \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$$2B=2×[−11​02​]

$$2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$$2B=[−22​04​]

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Calculate $A + 2B$A+2B:

$$A + 2B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$$A+2B=[−23​12​]+[−22​04​]

$$A + 2B = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$$A+2B=[−45​16​]

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Calculate $(A + 2B)^T$(A+2B)T:

$$(A + 2B)^T = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}^T$$(A+2B)T=[−45​16​]T

$$(A + 2B)^T = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$$(A+2B)T=[−41​56​]

Therefore, $(A + 2B)^T = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$(A+2B)T=[−41​56​]