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Answers without the blur. Sign up and see all textbooks for free! Q23.

Expert-verified Found in: Page 164 ### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903 # Perform the indicated matrix operation. If the matrix doesn’t exist, write impossible.23. $5\left[\begin{array}{ccc}\frac{1}{2}& 0& 1\\ 2& \frac{1}{3}& -1\end{array}\right]+4\left[\begin{array}{ccc}-2& \frac{3}{4}& 1\\ \frac{1}{6}& 0& \frac{5}{8}\end{array}\right]$

The final matrix after performing the indicated matrix operations is

$5\left[\begin{array}{ccc}\frac{1}{2}& 0& 1\\ 2& \frac{1}{3}& -1\end{array}\right]+4\left[\begin{array}{ccc}-2& \frac{3}{4}& 1\\ \frac{1}{6}& 0& \frac{5}{8}\end{array}\right]=\left[\begin{array}{ccc}\frac{-11}{2}& 3& 9\\ \frac{32}{3}& \frac{5}{3}& -\frac{5}{2}\end{array}\right]$.

See the step by step solution

## Step 1 - Define the concept used

If A and B are two matrices of order $m×n$, then the addition of the two matrices will also yield a matrix of order $m×n$ wherein each element is the sum of the corresponding elements.

The product of a scalar k and an $m×n$ matrix is an $m×n$ matrix in which each element equals k times the corresponding elements of the original matrix.

## Step 2 - Perform the scalar multiplication first

Multiply each element in the first matrix $\left[\begin{array}{ccc}\frac{1}{2}& 0& 1\\ 2& \frac{1}{3}& -1\end{array}\right]$ by 5 and multiply each element in the second matrix $\left[\begin{array}{ccc}-2& \frac{3}{4}& 1\\ \frac{1}{6}& 0& \frac{5}{8}\end{array}\right]$ by 4.

$\begin{array}{c}5\left[\begin{array}{ccc}\frac{1}{2}& 0& 1\\ 2& \frac{1}{3}& -1\end{array}\right]+4\left[\begin{array}{ccc}-2& \frac{3}{4}& 1\\ \frac{1}{6}& 0& \frac{5}{8}\end{array}\right]=\left[\begin{array}{ccc}5\left(\frac{1}{2}\right)& 5\left(0\right)& 5\left(1\right)\\ 5\left(2\right)& 5\left(\frac{1}{3}\right)& 5\left(-1\right)\end{array}\right]+\left[\begin{array}{ccc}4\left(-2\right)& 4\left(\frac{3}{4}\right)& 4\left(1\right)\\ 4\left(\frac{1}{6}\right)& 4\left(0\right)& 4\left(\frac{5}{8}\right)\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{5}{2}& 0& 5\\ 10& \frac{5}{3}& -5\end{array}\right]+\left[\begin{array}{ccc}-8& 3& 4\\ \frac{2}{3}& 0& \frac{5}{2}\end{array}\right]\end{array}$

## Step 3 - Add corresponding elements

Add corresponding elements of both the matrices $\left[\begin{array}{ccc}\frac{5}{2}& 0& 5\\ 10& \frac{5}{3}& -5\end{array}\right]+\left[\begin{array}{ccc}-8& 3& 4\\ \frac{2}{3}& 0& \frac{5}{2}\end{array}\right]$and simplify.

$\begin{array}{c}5\left[\begin{array}{ccc}\frac{1}{2}& 0& 1\\ 2& \frac{1}{3}& -1\end{array}\right]+4\left[\begin{array}{ccc}-2& \frac{3}{4}& 1\\ \frac{1}{6}& 0& \frac{5}{8}\end{array}\right]=\left[\begin{array}{ccc}\frac{5}{2}& 0& 5\\ 10& \frac{5}{3}& -5\end{array}\right]+\left[\begin{array}{ccc}-8& 3& 4\\ \frac{2}{3}& 0& \frac{5}{2}\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{5}{2}-8& 0+3& 5+4\\ 10+\frac{2}{3}& \frac{5}{3}+0& -5+\frac{5}{2}\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{-11}{2}& 3& 9\\ \frac{32}{3}& \frac{5}{3}& -\frac{5}{2}\end{array}\right]\end{array}$ ### Want to see more solutions like these? 