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 3end{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}2 58 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 3end{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}2 58 end{bmatrix}

Question
Matrices
asked 2020-10-28
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)

2020-10-29
Step 1
(a)\(x_1\begin{bmatrix}2 \\0 \end{bmatrix}+x_2\begin{bmatrix}1 \\3\end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
It is required to write the linear combination of columns
\(x_1\begin{bmatrix}2 \\0 \end{bmatrix}+x_2\begin{bmatrix}1 \\3\end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\) in a linear system.
Step 2
As per the definition of scalar multiplication:
\(x_1\begin{bmatrix}2 \\0 \end{bmatrix}=\begin{bmatrix}2x_1 \\0 \end{bmatrix}\)
\(x_2\begin{bmatrix}1 \\3 \end{bmatrix}=\begin{bmatrix}x_2 \\3x_2 \end{bmatrix}\)
So, \(\begin{bmatrix}2x_1 \\0 \end{bmatrix}+\begin{bmatrix}x_2 \\3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
Now by addition of matrices:
\(\begin{bmatrix}2x_1+x_2 \\0+3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
\(\begin{bmatrix}2x_1+x_2 \\3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}\)
Hence, the required linear system of given linear combination is:
\(2x_1+x_2=4\)
\(3x_2=2\)
Step 3
(b)\(x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}+x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}+x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}+x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}2 \\5\\8 \end{bmatrix}\)
It is required to write the linear combination of columns \(x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}+x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}+x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}+x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}2 \\5\\8 \end{bmatrix}\) in a linear system.
As per the definition of scalar multiplication:
\(x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}=\begin{bmatrix}x_1 \\2x_1\\-x_1 \end{bmatrix}\)
\(x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}=\begin{bmatrix}0 \\x_2\\2x_2 \end{bmatrix}\)
\(x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}=\begin{bmatrix}3x_3 \\4x_3\\5x_3 \end{bmatrix}\)
\(x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}x_4 \\3x_4\\4x_4 \end{bmatrix}\)
So, \(\begin{bmatrix}x_1 \\2x_1\\-x_1 \end{bmatrix}+\begin{bmatrix}0 \\x_2\\2x_2 \end{bmatrix}+\begin{bmatrix}3x_3 \\4x_3\\5x_3 \end{bmatrix}+\begin{bmatrix}x_4 \\3x_4\\4x_4 \end{bmatrix}=\begin{bmatrix}2 \\5\\8 \end{bmatrix}\)
Step 4
Now by addition of matrices:
\(\begin{bmatrix}x_1+0+3x_3+x_4\\2x_1+x_2+4x_3+3x_4\\-x_1+2x_2+5x_3+4x_4 \end{bmatrix}=\begin{bmatrix}2 \\5\\8 \end{bmatrix}\)
\(\begin{bmatrix}x_1+3x_3+x_4\\2x_1+x_2+4x_3+3x_4\\-x_1+2x_2+5x_3+4x_4 \end{bmatrix}=\begin{bmatrix}2 \\5\\8 \end{bmatrix}\)
Hence, the required linear system of given linear combination is:
\(x_1+3x_3+x_4=2\)
\(2x_1+x_2+4x_3+3x_4=5\)
\(-x_1+2x_2+5x_3+4x_4=8\)
0

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