Question

Given the matrices A and B shown below, find – B – 1/3A[ -18 3]. [ -4 12][ -15 -6] [ 8 -12]

Answer 1

[10 -13]

[-3 14]

Step-by-step explanation:

First, we will calculate 1/3A, so:

`(1)/(3)A=(1)/(3)\begin{bmatrix}{-18} & 3 & \\ {-15} & -6 & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{(1)/(3)(-18)} & {(1)/(3)(3)} & \\ {(1)/(3)(-15)} & {(1)/(3)(-6)} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-6} & {1} \\ {-5} & {-2} \\ & {}\end{bmatrix}`

Because 1/3 multiply each value in the matrix. Now, adding the respective values in the same position, we can calculate -B - 1/3A as:

`\begin{gathered} -B-(1)/(3)A=-\begin{bmatrix}{-4} & {12} & \\ {8} & {-12} & {}{}\end{bmatrix}-\begin{bmatrix}{-6} & {1} \\ {-5} & {-2}\end{bmatrix} \\ -B-(1)/(3)A=\begin{bmatrix}{4} & {-12} & \\ {-8} & {12} & {}{}\end{bmatrix}-\begin{bmatrix}{-6} & {1} \\ {-5} & {-2}\end{bmatrix} \\ -B-(1)/(3)A=\begin{bmatrix}{4-(-6)} & {-12-1} & \\ {-8-(-5)} & {12-(-2)} & {}{}\end{bmatrix} \\ -B-(1)/(3)A=\begin{bmatrix}{4+6} & {-12-1} & \\ {-8+5} & {12+2} & {}{}\end{bmatrix} \\ -B-(1)/(3)A=\begin{bmatrix}{10} & {-13} & \\ {-3} & {14} & {}{}\end{bmatrix} \end{gathered}`

Therefore, the answer is:

`-B-(1)/(3)A=\begin{bmatrix}{10} & {-13} & \\ {-3} & {14} & {}\end{bmatrix}`