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If \(\begin{bmatrix}2&-1\\[0.3em]2&0\end{bmatrix}\)+2A =\(\begin{bmatrix}-3&5\\[0.3em]4&3\end{bmatrix},\) the matrix A equals

(a) \(\begin{bmatrix}-5&6\\[0.3em]2&3\end{bmatrix}\)

(b) \(\begin{bmatrix}-\frac52&3\\[0.3em]1&\frac32\end{bmatrix}\)

(c) \(\begin{bmatrix}-\frac52&6\\[0.3em]2&3\end{bmatrix}\)

(d) \(\begin{bmatrix}-5&8\\[0.3em]1&3\end{bmatrix}\)

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(b) \(\begin{bmatrix}2&-1\\[0.3em]2&0\end{bmatrix}\) + 2A = \(\begin{bmatrix}-3&5\\[0.3em]4&3\end{bmatrix}\)

⇒ 2A = \(\begin{bmatrix}-3&5\\[0.3em]4&3\end{bmatrix}\) - \(\begin{bmatrix}2&-1\\[0.3em]2&0\end{bmatrix}\) - \(\begin{bmatrix}-3-2&5-(-1)\\[0.3em]4-2&3-0\end{bmatrix}\)

\(\begin{bmatrix}-5&6\\[0.3em]2&3\end{bmatrix}\)

= A = \(\frac12\)\(\begin{bmatrix}-5&6\\[0.3em]2&3\end{bmatrix}\) = \(\begin{bmatrix}-\frac52&3\\[0.3em]1&\frac32\end{bmatrix}\)

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