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

Given The Matrices A And B Shown Below Find B 13A 18 3 4 12 15 6 8 12 class=

Sagot :

Answer:

[10 -13]

[-3 14]

Explanation:

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

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

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:

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

Therefore, the answer is:

[tex]-B-\frac{1}{3}A=\begin{bmatrix}{10} & {-13} & \\ {-3} & {14} & {}\end{bmatrix}[/tex]

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