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2021-02-02

Find the following matrices: a) $A+B.$
(b) $A-B.$
(c) -4A.
(d) $3A+2B.$
$A=\left[\begin{array}{ccc}6& 2& -3\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 3\end{array}\right]$

Caren

Step 1
Given matrices:$A=\left[\begin{array}{ccc}6& 2& -3\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 3\end{array}\right]$
To find:
a) $A+B$.
(b) $A-B$.
(c) -4A.
(d) $3A+2B$.
Solution:
$A=\left[\begin{array}{ccc}6& 2& -3\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 3\end{array}\right]$
a)$A+B=\left[\begin{array}{ccc}6& 2& -3\end{array}\right]+\left[\begin{array}{ccc}4& -2& 3\end{array}\right]$
$⇒A+B=\left[\begin{array}{ccc}\left(6+4\right)& \left(2-2\right)& \left(-3+3\right)\end{array}\right]$
$⇒A+B=\left[\begin{array}{ccc}10& 0& 0\end{array}\right]$
b) $A-B=\left[\begin{array}{ccc}6& 2& -3\end{array}\right]-\left[\begin{array}{ccc}4& -2& 3\end{array}\right]$
$⇒A-B=\left[\begin{array}{ccc}\left(6-4\right)& \left(2-\left(-2\right)\right)& \left(-3-3\right)\end{array}\right]$
$⇒A-B=\left[\begin{array}{ccc}2& 4& -6\end{array}\right]$
c)$-4A=-4\left[\begin{array}{ccc}6& 2& -3\end{array}\right]$
$⇒-4A=\left[\begin{array}{ccc}-24& -8& 12\end{array}\right]$
d)$3A+2B=3\left[\begin{array}{ccc}6& 2& -3\end{array}\right]+2\left[\begin{array}{ccc}4& -2& 3\end{array}\right]$
$⇒3A+2B=\left[\begin{array}{ccc}18& 6& -9\end{array}\right]+\left[\begin{array}{ccc}8& -4& 6\end{array}\right]$
$⇒3A+2B=\left[\begin{array}{ccc}\left(18+8\right)& \left(6-4\right)& \left(-9+6\right)\end{array}\right]$
$⇒3A+2B=\left[\begin{array}{ccc}26& 2& -3\end{array}\right]$
Step 2
Result
a)$A+B=\left[\begin{array}{ccc}10& 0& 0\end{array}\right]$
b)$A-B=\left[\begin{array}{ccc}2& 4& -6\end{array}\right]$
c)$-4A=\left[\begin{array}{ccc}-24& -8& 12\end{array}\right]$
d)$3A+2B=\left[\begin{array}{ccc}26& 2& -3\end{array}\right]$

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