Question

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Answer 1

We conclude that:

`3\cdot \:2\begin{pmatrix}-2&3&0\\ -4&2&6\\ 6&-5&6\end{pmatrix}=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}`

Hence, option B is correct.

Explanation:

Given the expression

`3* 2\begin{pmatrix}-2&3&0\\ \:-4&2&6\\ \:6&-5&6\end{pmatrix}`

solving

`3* 2\begin{pmatrix}-2&3&0\\ \:-4&2&6\\ \:6&-5&6\end{pmatrix}`

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

`=\begin{pmatrix}3\cdot \:2\left(-2\right)&3\cdot \:2\cdot \:3&3\cdot \:2\cdot \:0\\ 3\cdot \:2\left(-4\right)&3\cdot \:2\cdot \:2&3\cdot \:2\cdot \:6\\ 3\cdot \:2\cdot \:6&3\cdot \:2\left(-5\right)&3\cdot \:2\cdot \:6\end{pmatrix}`

Simplify each element

`=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}`

Therefore, we conclude that:

`3\cdot \:2\begin{pmatrix}-2&3&0\\ -4&2&6\\ 6&-5&6\end{pmatrix}=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}`

Hence, option B is correct.