Amari Flowers

2021-03-11

Write each of the following linear combinations of columns as a linear system of the form in (4):
$a\right){x}_{1}\left[\begin{array}{c}2\\ 0\end{array}\right]+{x}_{2}\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]$
$b\right){x}_{1}\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]+{x}_{2}\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]+{x}_{3}\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]+{x}_{4}\left[\begin{array}{c}1\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]$

Yusuf Keller

Step 1
Given:
$a\right){x}_{1}\left[\begin{array}{c}2\\ 0\end{array}\right]+{x}_{2}\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]$
$b\right){x}_{1}\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]+{x}_{2}\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]+{x}_{3}\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]+{x}_{4}\left[\begin{array}{c}1\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]$
Step 2
$a\right){x}_{1}\left[\begin{array}{c}2\\ 0\end{array}\right]+{x}_{2}\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]$
Use the definition of scalar multiplication, we get
$\left[\begin{array}{c}2{x}_{1}\\ 0{x}_{1}\end{array}\right]+\left[\begin{array}{c}1{x}_{2}\\ 3{x}_{2}\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]$
$\left[\begin{array}{c}2{x}_{1}+{x}_{2}\\ 0{x}_{1}+3{x}_{2}\end{array}\right]=\left[\begin{array}{c}4\\ 2\end{array}\right]$
On comparing matrices, we get
$2{x}_{1}+{x}_{2}=4$
$3{x}_{2}=2$
Step 3
$b\right){x}_{1}\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]+{x}_{2}\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]+{x}_{3}\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]+{x}_{4}\left[\begin{array}{c}1\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]$
Using the definition of scalar mutiplication, we get
$\left[\begin{array}{c}1{x}_{1}\\ 2{x}_{1}\\ -1{x}_{1}\end{array}\right]+\left[\begin{array}{c}0{x}_{2}\\ 1{x}_{2}\\ 2{x}_{2}\end{array}\right]+\left[\begin{array}{c}3{x}_{3}\\ 4{x}_{3}\\ 5{x}_{3}\end{array}\right]+\left[\begin{array}{c}1{x}_{4}\\ 3{x}_{4}\\ 4{x}_{4}\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]$
Usinge addition of matrices, we get
$\left[\begin{array}{c}1{x}_{1}+0{x}_{2}+3{x}_{3}+1{x}_{4}\\ 2{x}_{1}+1{x}_{2}+4{x}_{3}+3{x}_{4}\\ -1{x}_{1}+2{x}_{2}+5{x}_{3}+4{x}_{4}\end{array}\right]=\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]$
On comparing matrices, we get
$1{x}_{1}+3{x}_{3}+1{x}_{4}=2$
$2{x}_{1}+1{x}_{2}+4{x}_{3}+3{x}_{4}=5$
$-1{x}_{1}+2{x}_{2}+5{x}_{3}+4{x}_{4}=8$

Jeffrey Jordon