Question

I have no idea what's going on

Answer 1

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

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}`