Question

I have no idea what's going on

Answer 1

Answer:

$3B-4C=\begin{pmatrix}-8&-23\\ 1&-15\end{pmatrix}$

Step-by-step explanation:

Let

$B=\begin{pmatrix}8&-5\\ -1&3\end{pmatrix}$

and

$C=\begin{pmatrix}8&2\\ \:-1&6\end{pmatrix}\:$

Finding 3B-4C

$\:3B-4C\:=\:3\begin{pmatrix}8&-5\\ -1&3\end{pmatrix}\:-4\begin{pmatrix}8&2\\ \:-1&6\end{pmatrix}$

first solving

$3\begin{pmatrix}8&-5\\ -1&3\end{pmatrix}$

Scalar multiplication: Multiply each of the matrix elements by a scalar

$3\begin{pmatrix}8&-5\\ \:\:-1&3\end{pmatrix}=\begin{pmatrix}3\cdot \:\:8&3\left(-5\right)\\ \:3\left(-1\right)&3\cdot \:\:3\end{pmatrix}$

simplify each element

                      $=\begin{pmatrix}24&-15\\ -3&9\end{pmatrix}$

now solving

$4\begin{pmatrix}8&2\\ \:\:-1&6\end{pmatrix}$

Scalar multiplication: Multiply each of the matrix elements by a scalar

$4\begin{pmatrix}8&2\\ \:\:-1&6\end{pmatrix}=\begin{pmatrix}4\cdot \:\:8&4\cdot \:\:2\\ \:4\left(-1\right)&4\cdot \:\:6\end{pmatrix}$

simplify each element

                    $=\begin{pmatrix}32&8\\ -4&24\end{pmatrix}$

now combining the results

$\:3B-4C\:=\:3\begin{pmatrix}8&-5\\ -1&3\end{pmatrix}\:-4\begin{pmatrix}8&2\\ \:-1&6\end{pmatrix}$

                $=\begin{pmatrix}24&-15\\ -3&9\end{pmatrix}-\begin{pmatrix}32&8\\ -4&24\end{pmatrix}$

subtract the elements in the matching positions

                 $=\begin{pmatrix}24-32&\left(-15\right)-8\\ \left(-3\right)-\left(-4\right)&9-24\end{pmatrix}$

simplifying the elements

                 $=\begin{pmatrix}-8&-23\\ 1&-15\end{pmatrix}$

Therefore,

$3B-4C=\begin{pmatrix}-8&-23\\ 1&-15\end{pmatrix}$