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}

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
asked 2020-11-12
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. \(\begin{bmatrix}1&3&0&-2&6\\0&0&1&4&7\\0&0&0&0&0\end{bmatrix}\)

Answers (1)

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: \(x=\begin{bmatrix}6\\0\\7\\0\end{bmatrix}+s\begin{bmatrix}-3\\1\\0\\0\end{bmatrix}+t\begin{bmatrix}2\\0\\-4\\1\end{bmatrix}\)
0

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