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

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
$$\begin{bmatrix}2x_1 \\0x_1 \end{bmatrix}+\begin{bmatrix}1x_2 \\3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}$$
$$\begin{bmatrix}2x_1+x_2 \\0x_1+3x_2 \end{bmatrix}=\begin{bmatrix}4 \\2 \end{bmatrix}$$
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
$$1x_1+3x_3+1x_4=2$$
$$2x_1+1x_2+4x_3+3x_4=5$$
$$-1x_1+2x_2+5x_3+4x_4=8$$

Relevant Questions

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)$$x_1\begin{bmatrix}1 \\2\\-1 \end{bmatrix}+x_2\begin{bmatrix}0 \\1\\2 \end{bmatrix}+x_3\begin{bmatrix}3 \\4\\5 \end{bmatrix}+x_4\begin{bmatrix}1 \\3\\4 \end{bmatrix}=\begin{bmatrix}2 \\5\\8 \end{bmatrix}$$
write B as a linear combination of the other matrices, if possible.
$$B=\begin{bmatrix}2 & 3 \\-4 & 2 \end{bmatrix} , A_1=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} , A_2=\begin{bmatrix}0 &-1 \\1 & 0 \end{bmatrix} , A_3=\begin{bmatrix}1 &1 \\0 & 1 \end{bmatrix}$$
Write the given matrix equation as a system of linear equations without matrices.
$$\begin{bmatrix}-1 & 0&1 \\ 0 & -1 &0 \\ 0&1&1 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}-4 \\ 2 \\ 4 \end{bmatrix}$$
Write the given matrix equation as a system of linear equations without matrices.
$$\begin{bmatrix}4 & -7 \\ 2 &-3 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}=\begin{bmatrix}-3 \\1 \end{bmatrix}$$
The row echelon form of a system of linear equations is given.
(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or $$x_1,x_2,x_3, x_4$$
(b) Determine whether the system is consistent. If it is consistent, give the solution.
$$\begin{matrix}1 & 0 & 2 & -1 \\ 0 & 1 & -4 & -2\\0&0&0&0&0 \end{matrix}$$
The row echelon form of a system of linear equations is given.
(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or $$x_1,x_2,x_3, x_4$$
(b) Determine whether the system is consistent. If it is consistent, give the solution.
$$\begin{matrix}1 & 0 & 3 & 0 &1 \\ 0 & 1 & 4 & 3&2\\0&0&1&2&3\\0&0&0&0&0 \end{matrix}$$
Write the matrix equation as a system of linear equations without matrices.
$$\begin{bmatrix}-1 & 0&1 \\0 & -1&0\\0&1&1 \end{bmatrix}\begin{bmatrix}x \\ y \\z \end{bmatrix}=\begin{bmatrix}-4 \\ 2\\4 \end{bmatrix}$$
$$B=\begin{bmatrix}2 & 5 \\0 & 3 \end{bmatrix} , A_1=\begin{bmatrix}1 & 2 \\-1 & 1 \end{bmatrix} , A_2=\begin{bmatrix}0 &1 \\2 & 1 \end{bmatrix}$$
$$\begin{bmatrix}2 & 0&-1 \\0 & 3&0\\1&1&0 \end{bmatrix}\begin{bmatrix}x \\ y \\z \end{bmatrix}=\begin{bmatrix}6 \\ 9\\5 \end{bmatrix}$$
$$A=\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix} , B=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix} , C=\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix} , D=\begin{bmatrix}2 & -3&4 \\-3& 1&-2 \end{bmatrix}$$