2021-10-15

Let $A=\left[\begin{array}{ccc}4& 0& 5\\ -1& 3& 2\end{array}\right]$
$B=\left[\begin{array}{ccc}1& 1& 1\\ 3& 5& 7\end{array}\right]$
$C=\left[\begin{array}{ccc}2& & -3\\ 0& & 1\end{array}\right]$
Find $3A–B;C\cdot B+A.$

Bentley Leach

$A=\left[\begin{array}{ccc}4& 0& 5\\ -1& 3& 2\end{array}\right]$
$B=\left[\begin{array}{ccc}1& 1& 1\\ 3& 5& 7\end{array}\right]$
$C=\left[\begin{array}{ccc}2& & -3\\ 0& & 1\end{array}\right]$
1) $\left(3A–B\right)=3×\left[\begin{array}{ccc}4& 0& 5\\ -1& 3& 2\end{array}\right]-\left[\begin{array}{ccc}1& 1& 1\\ 3& 5& 7\end{array}\right]=$
$=\left[\begin{array}{ccc}12& 0& 15\\ -3& 9& 6\end{array}\right]-\left[\begin{array}{ccc}1& 1& 1\\ 3& 5& 7\end{array}\right]=$
$=\left[\begin{array}{ccc}11& -1& 14\\ -6& 4& -1\end{array}\right]$
2) $\left(C\cdot \right)=\left[\begin{array}{ccc}2& & -3\\ 0& & 1\end{array}\right]\cdot =\left[\begin{array}{ccc}2& & 0\\ -3& & 1\end{array}\right]$
$\left(C\cdot B\right)=\left[\begin{array}{ccc}2& & 0\\ -3& & 1\end{array}\right]×\left[\begin{array}{ccc}1& 1& 1\\ 3& 5& 7\end{array}\right]=$
$=\left[\begin{array}{ccc}2& 2& 2\\ 0& 2& 4\end{array}\right]$
$\left(C\cdot B+A\right)=\left[\begin{array}{ccc}2& 2& 2\\ 0& 2& 4\end{array}\right]+\left[\begin{array}{ccc}4& 0& 5\\ -1& 3& 2\end{array}\right]=$
$=\left[\begin{array}{ccc}6& 2& 7\\ -1& 3& 6\end{array}\right]$

Do you have a similar question?