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Worked Examples · Example 6

Q.If A=(123−1)A=\begin{pmatrix}1&2\\3&-1\end{pmatrix} and B=(0−124)B=\begin{pmatrix}0&-1\\2&4\end{pmatrix}, find 2A−3B2A-3B.

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Step 1 — Scalar multiply each matrix separately

2A=(2(1)2(2)2(3)2(−1))=(246−2)2A = \begin{pmatrix}2(1)&2(2)\\2(3)&2(-1)\end{pmatrix} = \begin{pmatrix}2&4\\6&-2\end{pmatrix}

3B=(3(0)3(−1)3(2)3(4))=(0−3612)3B = \begin{pmatrix}3(0)&3(-1)\\3(2)&3(4)\end{pmatrix} = \begin{pmatrix}0&-3\\6&12\end{pmatrix}

Step 2 — Subtract element-by-element

2A−3B=(2−04−(−3)6−6−2−12)=(270−14)2A-3B = \begin{pmatrix}2-0&4-(-3)\\6-6&-2-12\end{pmatrix} = \begin{pmatrix}2&7\\0&-14\end{pmatrix} …

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