Question

Given the following matrices, if possible, determine 2A. if not, state “not possible”

Answer 1

`\begin{equation*} \begin{bmatrix}{-10} & {} & {0} \\ {8} & {} & {10} \\ {8} & {} & {6}\end{bmatrix} \end{equation*}`Step-by-step explanation

given

`A=\begin{bmatrix}{-5} & {} & {0} \\ {4} & {} & {5} \\ {4} & {} & {3}\end{bmatrix}`

Step 1

a) find 2A:

When you multiply a matrix by a number, you multiply every element in the matrix by the same number

`\begin{gathered} if \\ M=\begin{bmatrix}{a} & {b} \\ {c} & {d}\end{bmatrix} \\ then \\ kM=k\begin{bmatrix}{a} & {b} \\ {c} & {d}\end{bmatrix}=\begin{bmatrix}k{a} & {kb} \\ k{c} & k{d}\end{bmatrix} \end{gathered}`

hence

`\begin{gathered} A=\begin{bmatrix}{-5} & {} & {0} \\ {4} & {} & {5} \\ {4} & {} & {3}\end{bmatrix} \\ 2A=2\begin{bmatrix}{-5} & {} & {0} \\ {4} & {} & {5} \\ {4} & {} & {3}\end{bmatrix}=\begin{bmatrix}{-5*2} & {} & {0*2} \\ {4*2} & {} & {5*2} \\ {4*2} & {} & {3*2}\end{bmatrix} \\ 2A=\begin{bmatrix}{-10} & {} & {0} \\ {8} & {} & {10} \\ {8} & {} & {6}\end{bmatrix} \end{gathered}`

therefore, the answer is

`\begin{equation*} \begin{bmatrix}{-10} & {} & {0} \\ {8} & {} & {10} \\ {8} & {} & {6}\end{bmatrix} \end{equation*}`

I hope this helps you