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Exercise: Matrix Addition and Scalar ... · Q15

Q.For the matrices AA and BB of Q1, find 2A−3B2A-3B.

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2A=[4−206]2A=\begin{bmatrix}4&-2\\0&6\end{bmatrix} and 3B=[312−615]3B=\begin{bmatrix}3&12\\-6&15\end{bmatrix}. Subtracting entrywise: (1,1)(1,1): 4−3=14-3=1; (1,2)(1,2): −2−12=−14-2-12=-14; (2,1)(2,1): 0−(−6)=60-(-6)=6; (2,2)(2,2): 6−15=−96-15=-9. So 2A−3B=[1−146−9]2A-3B=\begin{bmatrix}1&-14\\6&-9\end{bmatrix}. [!ANSWER] 2A−3B=[1−146−9]2A-3B=\begin{bmatrix}1&-14\\6&-9\end{bmatrix}.

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