Math Problem Solver

The system of equations has one solution.\[x = \frac{1}{13}\] \[y = \frac{5}{8}\] \[z = \frac{31}{104}\]

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}\frac{3}{5} & - \frac{1}{20} & - \frac{1}{20} &\bigm |& 0\\- \frac{1}{2} & \frac{3}{10} & - \frac{1}{2} &\bigm |& 0\\- \frac{1}{10} & - \frac{1}{4} & \frac{11}{20} &\bigm |& 0\\1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\- \frac{1}{2} & \frac{3}{10} & - \frac{1}{2} &\bigm |& 0\\- \frac{1}{10} & - \frac{1}{4} & \frac{11}{20} &\bigm |& 0\\\frac{3}{5} & - \frac{1}{20} & - \frac{1}{20} &\bigm |& 0\end{bmatrix}$$

The leading term of row \(1\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\- \frac{1}{2} & \frac{3}{10} & - \frac{1}{2} &\bigm |& 0\\- \frac{1}{10} & - \frac{1}{4} & \frac{11}{20} &\bigm |& 0\\\frac{3}{5} & - \frac{1}{20} & - \frac{1}{20} &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(\frac{1}{2}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{4}{5} & 0 &\bigm |& \frac{1}{2}\\- \frac{1}{10} & - \frac{1}{4} & \frac{11}{20} &\bigm |& 0\\\frac{3}{5} & - \frac{1}{20} & - \frac{1}{20} &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(\frac{1}{10}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{4}{5} & 0 &\bigm |& \frac{1}{2}\\0 & - \frac{3}{20} & \frac{13}{20} &\bigm |& \frac{1}{10}\\\frac{3}{5} & - \frac{1}{20} & - \frac{1}{20} &\bigm |& 0\end{bmatrix}$$

Multiply row \(1\) by scalar \(- \frac{3}{5}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{4}{5} & 0 &\bigm |& \frac{1}{2}\\0 & - \frac{3}{20} & \frac{13}{20} &\bigm |& \frac{1}{10}\\0 & - \frac{13}{20} & - \frac{13}{20} &\bigm |& - \frac{3}{5}\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & \frac{4}{5} & 0 &\bigm |& \frac{1}{2}\\0 & - \frac{3}{20} & \frac{13}{20} &\bigm |& \frac{1}{10}\\0 & - \frac{13}{20} & - \frac{13}{20} &\bigm |& - \frac{3}{5}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{5}{4}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 1 & 1 &\bigm |& 1\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & - \frac{3}{20} & \frac{13}{20} &\bigm |& \frac{1}{10}\\0 & - \frac{13}{20} & - \frac{13}{20} &\bigm |& - \frac{3}{5}\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& \frac{3}{8}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & - \frac{3}{20} & \frac{13}{20} &\bigm |& \frac{1}{10}\\0 & - \frac{13}{20} & - \frac{13}{20} &\bigm |& - \frac{3}{5}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{3}{20}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& \frac{3}{8}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & 0 & \frac{13}{20} &\bigm |& \frac{31}{160}\\0 & - \frac{13}{20} & - \frac{13}{20} &\bigm |& - \frac{3}{5}\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{13}{20}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& \frac{3}{8}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & 0 & \frac{13}{20} &\bigm |& \frac{31}{160}\\0 & 0 & - \frac{13}{20} &\bigm |& - \frac{31}{160}\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& \frac{3}{8}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & 0 & \frac{13}{20} &\bigm |& \frac{31}{160}\\0 & 0 & - \frac{13}{20} &\bigm |& - \frac{31}{160}\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{20}{13}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & 1 &\bigm |& \frac{3}{8}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & 0 & 1 &\bigm |& \frac{31}{104}\\0 & 0 & - \frac{13}{20} &\bigm |& - \frac{31}{160}\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{1}{13}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & 0 & 1 &\bigm |& \frac{31}{104}\\0 & 0 & - \frac{13}{20} &\bigm |& - \frac{31}{160}\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{13}{20}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{1}{13}\\0 & 1 & 0 &\bigm |& \frac{5}{8}\\0 & 0 & 1 &\bigm |& \frac{31}{104}\\0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{1}{13} \\ x = \frac{1}{13}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 1 \cdot y+ 0 \cdot z = \frac{5}{8} \\ y = \frac{5}{8}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 0 \cdot y+ 1 \cdot z = \frac{31}{104} \\ z = \frac{31}{104}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 0 \cdot y+ 0 \cdot z = 0 \\ None = 0\end{aligned}$$