Math Problem Solver

The system of equations has one solution.\[w = 0\] \[x = - \frac{280}{3}\] \[y = \frac{2135}{9}\] \[z = - \frac{665}{9}\]

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}0 & 1 & 1 & 1 &\bigm |& 70\\11 & 0 & 5 & 8 &\bigm |& 595\\6 & 5 & 3 & 0 &\bigm |& 245\\-3 & 8 & 3 & 0 &\bigm |& -35\\0 & 11 & 6 & 3 &\bigm |& 175\end{bmatrix}$$

Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:

  1. Swap the positions of any two rows.

  2. Multiply any row by a nonzero scalar.

  3. Multiply a row by a nonzero scalar and add it to any other row.

First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.

$$\begin{bmatrix}11 & 0 & 5 & 8 &\bigm |& 595\\0 & 1 & 1 & 1 &\bigm |& 70\\6 & 5 & 3 & 0 &\bigm |& 245\\-3 & 8 & 3 & 0 &\bigm |& -35\\0 & 11 & 6 & 3 &\bigm |& 175\end{bmatrix}$$

Multiply row \(1\) by scalar \(\frac{1}{11}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & 1 & 1 &\bigm |& 70\\6 & 5 & 3 & 0 &\bigm |& 245\\-3 & 8 & 3 & 0 &\bigm |& -35\\0 & 11 & 6 & 3 &\bigm |& 175\end{bmatrix}$$

Multiply row \(1\) by scalar \(-6\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & 1 & 1 &\bigm |& 70\\0 & 5 & \frac{3}{11} & - \frac{48}{11} &\bigm |& - \frac{875}{11}\\-3 & 8 & 3 & 0 &\bigm |& -35\\0 & 11 & 6 & 3 &\bigm |& 175\end{bmatrix}$$

Multiply row \(1\) by scalar \(3\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & 1 & 1 &\bigm |& 70\\0 & 5 & \frac{3}{11} & - \frac{48}{11} &\bigm |& - \frac{875}{11}\\0 & 8 & \frac{48}{11} & \frac{24}{11} &\bigm |& \frac{1400}{11}\\0 & 11 & 6 & 3 &\bigm |& 175\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 11 & 6 & 3 &\bigm |& 175\\0 & 5 & \frac{3}{11} & - \frac{48}{11} &\bigm |& - \frac{875}{11}\\0 & 8 & \frac{48}{11} & \frac{24}{11} &\bigm |& \frac{1400}{11}\\0 & 1 & 1 & 1 &\bigm |& 70\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{1}{11}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 5 & \frac{3}{11} & - \frac{48}{11} &\bigm |& - \frac{875}{11}\\0 & 8 & \frac{48}{11} & \frac{24}{11} &\bigm |& \frac{1400}{11}\\0 & 1 & 1 & 1 &\bigm |& 70\end{bmatrix}$$

Multiply row \(2\) by scalar \(-5\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 0 & - \frac{27}{11} & - \frac{63}{11} &\bigm |& - \frac{1750}{11}\\0 & 8 & \frac{48}{11} & \frac{24}{11} &\bigm |& \frac{1400}{11}\\0 & 1 & 1 & 1 &\bigm |& 70\end{bmatrix}$$

Multiply row \(2\) by scalar \(-8\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 0 & - \frac{27}{11} & - \frac{63}{11} &\bigm |& - \frac{1750}{11}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 1 & 1 & 1 &\bigm |& 70\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 0 & - \frac{27}{11} & - \frac{63}{11} &\bigm |& - \frac{1750}{11}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 0 & - \frac{27}{11} & - \frac{63}{11} &\bigm |& - \frac{1750}{11}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{11}{27}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{5}{11}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\\0 & 1 & \frac{6}{11} & \frac{3}{11} &\bigm |& \frac{175}{11}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{6}{11}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\\0 & 1 & 0 & -1 &\bigm |& - \frac{175}{9}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & \frac{5}{11} & \frac{8}{11} &\bigm |& \frac{595}{11}\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{5}{11}\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\\0 & 1 & 0 & -1 &\bigm |& - \frac{175}{9}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\end{bmatrix}$$

Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.

$$\begin{bmatrix}1 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\\0 & 1 & 0 & -1 &\bigm |& - \frac{175}{9}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(4\) by scalar \(-3\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & 0 & - \frac{1}{3} &\bigm |& \frac{665}{27}\\0 & 1 & 0 & -1 &\bigm |& - \frac{175}{9}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 1 &\bigm |& - \frac{665}{9}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(4\) by scalar \(\frac{1}{3}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& 0\\0 & 1 & 0 & -1 &\bigm |& - \frac{175}{9}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 1 &\bigm |& - \frac{665}{9}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(4\) by scalar \(1\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& 0\\0 & 1 & 0 & 0 &\bigm |& - \frac{280}{3}\\0 & 0 & 1 & \frac{7}{3} &\bigm |& \frac{1750}{27}\\0 & 0 & 0 & 1 &\bigm |& - \frac{665}{9}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{7}{3}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& 0\\0 & 1 & 0 & 0 &\bigm |& - \frac{280}{3}\\0 & 0 & 1 & 0 &\bigm |& \frac{2135}{9}\\0 & 0 & 0 & 1 &\bigm |& - \frac{665}{9}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot w+ 0 \cdot x+ 0 \cdot y+ 0 \cdot z = 0 \\ w = 0\end{aligned}$$$$\begin{aligned}0 \cdot w+ 1 \cdot x+ 0 \cdot y+ 0 \cdot z = - \frac{280}{3} \\ x = - \frac{280}{3}\end{aligned}$$$$\begin{aligned}0 \cdot w+ 0 \cdot x+ 1 \cdot y+ 0 \cdot z = \frac{2135}{9} \\ y = \frac{2135}{9}\end{aligned}$$$$\begin{aligned}0 \cdot w+ 0 \cdot x+ 0 \cdot y+ 1 \cdot z = - \frac{665}{9} \\ z = - \frac{665}{9}\end{aligned}$$$$\begin{aligned}0 \cdot w+ 0 \cdot x+ 0 \cdot y+ 0 \cdot z = 0 \\ None = 0\end{aligned}$$