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Mathematics

If [23−41]+2A=4[−1234]\begin{bmatrix} 2 & 3 \ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \ 3 & 4 \end{bmatrix}, then the matrix A is:

  1. [−35815]\begin{bmatrix} -3 & 5 \ 8 & 15 \end{bmatrix}

  2. [5−3158]\begin{bmatrix} 5 & -3 \ 15 & 8 \end{bmatrix}

  3. [−3528152]\begin{bmatrix} -3 & \dfrac{5}{2} \ 8 & \dfrac{15}{2} \end{bmatrix}

  4. [52−31528]\begin{bmatrix} \dfrac{5}{2} & -3 \ \dfrac{15}{2} & 8 \end{bmatrix}

Matrices

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Answer

Given,

⇒[23−41]+2A=4[−1234]⇒2A=4[−1234]−[23−41]⇒A=12(4[−1234]−[23−41])⇒A=12([−481216]−[23−41])⇒A=12([−4−28−312−(−4)16−1])⇒12[−651615]⇒[−3528152].\Rightarrow \begin{bmatrix} 2 & 3 \ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow 2A = 4 \begin{bmatrix} -1 & 2 \ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \ -4 & 1 \end{bmatrix} \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(4\begin{bmatrix} -1 & 2 \ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(\begin{bmatrix} -4 & 8 \ 12 & 16 \end{bmatrix} - \begin{bmatrix} 2 & 3 \ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \dfrac{1}{2} \Big(\begin{bmatrix} -4 - 2 & 8 - 3 \ 12 - (-4) & 16 - 1 \end{bmatrix}\Big) \\[1em] \Rightarrow \dfrac{1}{2}\begin{bmatrix} -6 & 5 \ 16 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 & \dfrac{5}{2} \\ 8 & \dfrac{15}{2} \end{bmatrix}.

Hence, option 3 is the correct option.

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