Ayaana Buck

2021-03-07

Find the following matrices:
a) $A+B$.
(b) $A-B$.
(c) $-4A$.
$A=\left[\begin{array}{ccc}2& -10& -2\\ 14& 12& 10\\ 4& -2& 2\end{array}\right],B=\left[\begin{array}{ccc}6& 10& -2\\ 0& -12& -4\\ -5& 2& -2\end{array}\right]$

Elberte

Given,
$A=\left[\begin{array}{ccc}2& -10& -2\\ 14& 12& 10\\ 4& -2& 2\end{array}\right],B=\left[\begin{array}{ccc}6& 10& -2\\ 0& -12& -4\\ -5& 2& -2\end{array}\right]$
a)$A+B=\left[\begin{array}{ccc}2& -10& -2\\ 14& 12& 10\\ 4& -2& 2\end{array}\right]+\left[\begin{array}{ccc}6& 10& -2\\ 0& -12& -4\\ -5& 2& -2\end{array}\right]$
$=\left[\begin{array}{ccc}2+6& -10+10& -2-2\\ 14+0& 12-12& 10-4\\ 4-5& -2+2& 2-2\end{array}\right]$
$=\left[\begin{array}{ccc}8& 0& -4\\ 14& 0& 6\\ -1& 0& 0\end{array}\right]$
Step 2
b)$A-B=\left[\begin{array}{ccc}2& -10& -2\\ 14& 12& 10\\ 4& -2& 2\end{array}\right]-\left[\begin{array}{ccc}6& 10& -2\\ 0& -12& -4\\ -5& 2& -2\end{array}\right]$
$=\left[\begin{array}{ccc}2-6& -10-10& -2-\left(-2\right)\\ 14-0& 12-\left(-12\right)& 10-\left(-4\right)\\ 4-\left(-5\right)& -2-2& 2-\left(-2\right)\end{array}\right]$
$=\left[\begin{array}{ccc}-4& -20& 0\\ 14& 24& 14\\ 9& -4& 4\end{array}\right]$
c)$-4A=-4\left[\begin{array}{ccc}2& -10& -2\\ 14& 12& 10\\ 4& -2& 2\end{array}\right]$
$=\left[\begin{array}{ccc}-4\left(2\right)& -4\left(-10\right)& -4\left(-2\right)\\ -4\left(14\right)& -4\left(12\right)& -4\left(10\right)\\ -4\left(4\right)& -4\left(-2\right)& -4\left(2\right)\end{array}\right]$
$=\left[\begin{array}{ccc}-8& 40& 8\\ -56& -48& -40\\ -16& 8& -8\end{array}\right]$

Jeffrey Jordon

Answer is given below (on video)

Jeffrey Jordon