Ernstfalld

2020-11-14

Multiply the Following matrices:
$\left[\begin{array}{cc}2& 3\\ 1& 0\end{array}\right]×\left[\begin{array}{cc}4& 1\\ 2& 1\end{array}\right]$
$\left[\begin{array}{cc}1& 3\\ 4& 5\\ 1& 2\end{array}\right]×\left[\begin{array}{ccc}2& -1& 4\\ 3& 1& 0\end{array}\right]$

Demi-Leigh Barrera

Step 1
Multiply the first set of matrices as,
$=\left[\begin{array}{cc}2& 3\\ 1& 0\end{array}\right]×\left[\begin{array}{cc}4& 1\\ 2& 1\end{array}\right]$
$=\left[\begin{array}{cc}2& 3\\ 1& 0\end{array}\right]×\left[\begin{array}{cc}4& 1\\ 2& 1\end{array}\right]$
$=\left[\begin{array}{cc}2\left(4\right)+3\left(2\right)& 2\left(1\right)+3\left(1\right)\\ 1\left(4\right)+0\left(2\right)& 1\left(1\right)+0\left(1\right)\end{array}\right]$
$=\left[\begin{array}{cc}8+6& 2+3\\ 4+0& 1+0\end{array}\right]$
$=\left[\begin{array}{cc}14& 5\\ 4& 1\end{array}\right]$
Thus, the multiplication gives $\left[\begin{array}{cc}2& 3\\ 1& 0\end{array}\right]×\left[\begin{array}{cc}4& 1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}14& 5\\ 4& 1\end{array}\right]$
Step 2
Multiply the second set of matrices as,
$\left[\begin{array}{cc}1& 3\\ 4& 5\\ 1& 2\end{array}\right]×\left[\begin{array}{ccc}2& -1& 4\\ 3& 1& 0\end{array}\right]$
$\left[\begin{array}{cc}1& 3\\ 4& 5\\ 1& 2\end{array}\right]×\left[\begin{array}{ccc}2& -1& 4\\ 3& 1& 0\end{array}\right]$
$=\left[\begin{array}{ccc}1\left(2\right)+3\left(3\right)& 1\left(-1\right)+3\left(1\right)& 1\left(4\right)+3\left(0\right)\\ 4\left(2\right)+5\left(3\right)& 4\left(-1\right)+5\left(1\right)& 4\left(4\right)+5\left(0\right)\\ 1\left(2\right)+2\left(3\right)& 1\left(-1\right)+2\left(1\right)& 1\left(4\right)+2\left(0\right)\end{array}\right]$
$=\left[\begin{array}{ccc}2+9& -1+3& 4+0\\ 8+15& -4+5& 16+0\\ 2+6& -1+2& 4+0\end{array}\right]$
$=\left[\begin{array}{ccc}11& 2& 4\\ 23& 1& 16\\ 8& 1& 4\end{array}\right]$
Thus, the multiplication gives $\left[\begin{array}{cc}1& 3\\ 4& 5\\ 1& 2\end{array}\right]×\left[\begin{array}{ccc}2& -1& 4\\ 3& 1& 0\end{array}\right]=\left[\begin{array}{ccc}11& 2& 4\\ 23& 1& 16\\ 8& 1& 4\end{array}\right]$

Jeffrey Jordon