Write each of the following linear combinations of columns as a linear system of the form in (4): a) x_1begin{bmatrix}2 0 end{bmatrix}+x_2begin{bmatrix}1 3 end{bmatrix}=begin{bmatrix}4 2 end{bmatrix} b) x_1begin{bmatrix}1 2-1 end{bmatrix}+x_2begin{bmatrix}0 12 end{bmatrix}+x_3begin{bmatrix}3 45 end{bmatrix}+x_4begin{bmatrix}1 34 end{bmatrix}=begin{bmatrix}258 end{bmatrix}

Write each of the following linear combinations of columns as a linear system of the form in (4): a) x_1begin{bmatrix}2 0 end{bmatrix}+x_2begin{bmatrix}1 3 end{bmatrix}=begin{bmatrix}4 2 end{bmatrix} b) x_1begin{bmatrix}1 2-1 end{bmatrix}+x_2begin{bmatrix}0 12 end{bmatrix}+x_3begin{bmatrix}3 45 end{bmatrix}+x_4begin{bmatrix}1 34 end{bmatrix}=begin{bmatrix}258 end{bmatrix}

Question
Matrices
asked 2021-03-11
Write each of the following linear combinations of columns as a linear system of the form in (4):
\(a) x_1\begin{bmatrix}2 \\0 \end{bmatrix}+x_2\begin{bmatrix}1 \\3 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
\(b) x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}+x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}+x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}+x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}2\\5\\8 \end{bmatrix}\)

Answers (1)

2021-03-12
Step 1
Given:
\(a) x_1\begin{bmatrix}2 \\0 \end{bmatrix}+x_2\begin{bmatrix}1 \\3 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
\(b) x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}+x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}+x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}+x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}2\\5\\8 \end{bmatrix}\)
Step 2
\(a) x_1\begin{bmatrix}2 \\0 \end{bmatrix}+x_2\begin{bmatrix}1 \\3 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
Use the definition of scalar multiplication, we get
\(\begin{bmatrix}2x_1 \\0x_1 \end{bmatrix}+\begin{bmatrix}1x_2 \\3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
Using addition of matrix
\(\begin{bmatrix}2x_1+x_2 \\0x_1+3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
On comparing matrices, we get
\(2x_1+x_2=4\)
\(3x_2=2\)
Step 3
\(b) x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}+x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}+x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}+x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}2\\5\\8 \end{bmatrix}\)
Using the definition of scalar mutiplication, we get
\(\begin{bmatrix}1x_1 \\2x_1\\-1x_1 \end{bmatrix}+\begin{bmatrix}0x_2 \\1x_2\\2x_2 \end{bmatrix}+\begin{bmatrix}3x_3 \\4x_3\\5x_3 \end{bmatrix}+\begin{bmatrix}1x_4 \\3x_4\\4x_4 \end{bmatrix}=\begin{bmatrix}2\\5\\8 \end{bmatrix}\)
Usinge addition of matrices, we get
\(\begin{bmatrix}1x_1+0x_2+3x_3+1x_4 \\2x_1+1x_2+4x_3+3x_4\\-1x_1+2x_2+5x_3+4x_4 \end{bmatrix}=\begin{bmatrix}2\\5\\8 \end{bmatrix}\)
On comparing matrices, we get
\(1x_1+3x_3+1x_4=2\)
\(2x_1+1x_2+4x_3+3x_4=5\)
\(-1x_1+2x_2+5x_3+4x_4=8\)
0

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