Math Problem Solver

The system of equations has one solution.\[t = \frac{18}{149}\] \[x = \frac{50}{149}\] \[y = \frac{45}{149}\] \[z = \frac{36}{149}\]

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}-1 & \frac{9}{10} & - \frac{1}{5} & - \frac{1}{2} &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\1 & 0 & 0 & - \frac{1}{2} &\bigm |& 0\\1 & 1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

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.

$$\begin{bmatrix}-1 & \frac{9}{10} & - \frac{1}{5} & - \frac{1}{2} &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\1 & 0 & 0 & - \frac{1}{2} &\bigm |& 0\\1 & 1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

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

$$\begin{bmatrix}1 & - \frac{9}{10} & \frac{1}{5} & \frac{1}{2} &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\1 & 0 & 0 & - \frac{1}{2} &\bigm |& 0\\1 & 1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

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

$$\begin{bmatrix}1 & - \frac{9}{10} & \frac{1}{5} & \frac{1}{2} &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & \frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\\1 & 1 & 1 & 1 &\bigm |& 1\end{bmatrix}$$

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

$$\begin{bmatrix}1 & - \frac{9}{10} & \frac{1}{5} & \frac{1}{2} &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & \frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\\0 & \frac{19}{10} & \frac{4}{5} & \frac{1}{2} &\bigm |& 1\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & - \frac{9}{10} & \frac{1}{5} & \frac{1}{2} &\bigm |& 0\\0 & \frac{19}{10} & \frac{4}{5} & \frac{1}{2} &\bigm |& 1\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & \frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{10}{19}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & - \frac{9}{10} & \frac{1}{5} & \frac{1}{2} &\bigm |& 0\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & \frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{9}{10}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & \frac{11}{19} & \frac{14}{19} &\bigm |& \frac{9}{19}\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & \frac{9}{10} & - \frac{1}{5} & -1 &\bigm |& 0\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(- \frac{9}{10}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & \frac{11}{19} & \frac{14}{19} &\bigm |& \frac{9}{19}\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & 0 & - \frac{11}{19} & - \frac{47}{38} &\bigm |& - \frac{9}{19}\\0 & - \frac{9}{10} & 1 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(2\) by scalar \(\frac{9}{10}\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & \frac{11}{19} & \frac{14}{19} &\bigm |& \frac{9}{19}\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\\0 & 0 & - \frac{11}{19} & - \frac{47}{38} &\bigm |& - \frac{9}{19}\\0 & 0 & \frac{131}{95} & \frac{9}{38} &\bigm |& \frac{9}{19}\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & \frac{11}{19} & \frac{14}{19} &\bigm |& \frac{9}{19}\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & \frac{131}{95} & \frac{9}{38} &\bigm |& \frac{9}{19}\\0 & 0 & - \frac{11}{19} & - \frac{47}{38} &\bigm |& - \frac{9}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{95}{131}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & \frac{11}{19} & \frac{14}{19} &\bigm |& \frac{9}{19}\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & - \frac{11}{19} & - \frac{47}{38} &\bigm |& - \frac{9}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{11}{19}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{167}{262} &\bigm |& \frac{36}{131}\\0 & 1 & \frac{8}{19} & \frac{5}{19} &\bigm |& \frac{10}{19}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & - \frac{11}{19} & - \frac{47}{38} &\bigm |& - \frac{9}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(- \frac{8}{19}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{167}{262} &\bigm |& \frac{36}{131}\\0 & 1 & 0 & \frac{25}{131} &\bigm |& \frac{50}{131}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & - \frac{11}{19} & - \frac{47}{38} &\bigm |& - \frac{9}{19}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\end{bmatrix}$$

Multiply row \(3\) by scalar \(\frac{11}{19}\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{167}{262} &\bigm |& \frac{36}{131}\\0 & 1 & 0 & \frac{25}{131} &\bigm |& \frac{50}{131}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & 0 & - \frac{149}{131} &\bigm |& - \frac{36}{131}\\0 & 0 & - \frac{4}{5} & 1 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 0 & \frac{167}{262} &\bigm |& \frac{36}{131}\\0 & 1 & 0 & \frac{25}{131} &\bigm |& \frac{50}{131}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & 0 & - \frac{149}{131} &\bigm |& - \frac{36}{131}\\0 & 0 & 0 & \frac{149}{131} &\bigm |& \frac{36}{131}\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 0 & \frac{167}{262} &\bigm |& \frac{36}{131}\\0 & 1 & 0 & \frac{25}{131} &\bigm |& \frac{50}{131}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & 0 & - \frac{149}{131} &\bigm |& - \frac{36}{131}\\0 & 0 & 0 & \frac{149}{131} &\bigm |& \frac{36}{131}\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{131}{149}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 0 & 0 & \frac{167}{262} &\bigm |& \frac{36}{131}\\0 & 1 & 0 & \frac{25}{131} &\bigm |& \frac{50}{131}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & 0 & 1 &\bigm |& \frac{36}{149}\\0 & 0 & 0 & \frac{149}{131} &\bigm |& \frac{36}{131}\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{167}{262}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{18}{149}\\0 & 1 & 0 & \frac{25}{131} &\bigm |& \frac{50}{131}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & 0 & 1 &\bigm |& \frac{36}{149}\\0 & 0 & 0 & \frac{149}{131} &\bigm |& \frac{36}{131}\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{25}{131}\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{18}{149}\\0 & 1 & 0 & 0 &\bigm |& \frac{50}{149}\\0 & 0 & 1 & \frac{45}{262} &\bigm |& \frac{45}{131}\\0 & 0 & 0 & 1 &\bigm |& \frac{36}{149}\\0 & 0 & 0 & \frac{149}{131} &\bigm |& \frac{36}{131}\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{45}{262}\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{18}{149}\\0 & 1 & 0 & 0 &\bigm |& \frac{50}{149}\\0 & 0 & 1 & 0 &\bigm |& \frac{45}{149}\\0 & 0 & 0 & 1 &\bigm |& \frac{36}{149}\\0 & 0 & 0 & \frac{149}{131} &\bigm |& \frac{36}{131}\end{bmatrix}$$

Multiply row \(4\) by scalar \(- \frac{149}{131}\) and add it to row \(5\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& \frac{18}{149}\\0 & 1 & 0 & 0 &\bigm |& \frac{50}{149}\\0 & 0 & 1 & 0 &\bigm |& \frac{45}{149}\\0 & 0 & 0 & 1 &\bigm |& \frac{36}{149}\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot t+ 0 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{18}{149} \\ t = \frac{18}{149}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 1 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{50}{149} \\ x = \frac{50}{149}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 0 \cdot x+ 1 \cdot y+ 0 \cdot z = \frac{45}{149} \\ y = \frac{45}{149}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 0 \cdot x+ 0 \cdot y+ 1 \cdot z = \frac{36}{149} \\ z = \frac{36}{149}\end{aligned}$$$$\begin{aligned}0 \cdot t+ 0 \cdot x+ 0 \cdot y+ 0 \cdot z = 0 \\ None = 0\end{aligned}$$