The reduced row echelon form of the augmented matrix of a system of linear equations is given. Determine whether this system of linear equations is consistent and, if so, find its general solution. Write the solution in vector form. begin{bmatrix}1&3&0&-2&60&0&1&4&70&0&0&0&0end{bmatrix}

Question
Matrices
The reduced row echelon form of the augmented matrix of a system of linear equations is given. Determine whether this system of linear equations is consistent and, if so, find its general solution. Write the solution in vector form. $$\begin{bmatrix}1&3&0&-2&6\\0&0&1&4&7\\0&0&0&0&0\end{bmatrix}$$

2020-11-13
The augmented matrix does not contain a row in which the only nonzero entry appears in the last column. Therefore, this system of equations must be consistent. Convert the augmented matrix into a system of equantions.
$$\begin{cases}x_1+3x_2-2x_4=6\\x_3+4x_4=7\\0=0\end{cases}$$
Solve for the leading entry for each individual equation. Determine thee free variables, if any. $$x_1=6-3x_2+2x_4$$
$$x_2,\ free$$
$$x_3=7-4x_4$$
$$x_4,\ free$$
Parameterize the free variables.
$$\begin{cases}x_1=6-3s+2t\\x_2=s\\x_3=7-4t\\x_4=t\end{cases}$$
Write the solution in vector form.
$$x=\begin{bmatrix}6\\0\\7\\0\end{bmatrix}+s\begin{bmatrix}-3\\1\\0\\0\end{bmatrix}+t\begin{bmatrix}2\\0\\-4\\1\end{bmatrix}$$
Result:
consistent: $$x=\begin{bmatrix}6\\0\\7\\0\end{bmatrix}+s\begin{bmatrix}-3\\1\\0\\0\end{bmatrix}+t\begin{bmatrix}2\\0\\-4\\1\end{bmatrix}$$

Relevant Questions

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Use x, y, or x, y, z, or $$x_1,x_2,x_3, x_4$$
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(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or $$x_1,x_2,x_3, x_4$$
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$$\begin{matrix}1 & 0 & 3 & 0 &1 \\ 0 & 1 & 4 & 3&2\\0&0&1&2&3\\0&0&0&0&0 \end{matrix}$$
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$$\begin{bmatrix}1 & 0&-4&|&0 \\0&1&2&|&0\\0&0&0&|&1 \end{bmatrix}$$
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