# 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&-2&0&0&-30&0&1&0&-40&0&0&1&5end{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. $\left[\begin{array}{ccccc}1& -2& 0& 0& -3\\ 0& 0& 1& 0& -4\\ 0& 0& 0& 1& 5\end{array}\right]$
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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. $\left\{\begin{array}{l}{x}_{1}-2{x}_{2}=-3\\ {x}_{3}=-4\\ {x}_{4}=5\end{array}$
Solve for the leading entry for each individual equation. Determine thee free variables, if any. ${x}_{1}=-3+2{x}_{2}$

${x}_{3}=-4$
${x}_{4}=5$
Parameterize the free variables. $\left\{\begin{array}{l}{x}_{1}=-3+2{x}_{2}\\ {x}_{2}=t\\ {x}_{3}=-4\\ {x}_{4}=5\end{array}$
Write the solution in vector form. $x=\left[\begin{array}{c}-3\\ 0\\ -4\\ 5\end{array}\right]+t\left[\begin{array}{c}2\\ 1\\ 0\\ 0\end{array}\right]$
Result: consistent: $x=\left[\begin{array}{c}-3\\ 0\\ -4\\ 5\end{array}\right]+t\left[\begin{array}{c}2\\ 1\\ 0\\ 0\end{array}\right]$