Math Problem Solver

The system of equations has one solution.\[x = \frac{10}{19}\] \[y = \frac{9}{19}\]

Solve by Gauss-Jordan Elimination:

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

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.

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

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

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

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

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

Multiply row \(1\) by scalar \(- \frac{9}{10}\) and add it to row \(3\).

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

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.

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

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

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

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

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

Multiply row \(2\) by scalar \(\frac{19}{10}\) and add it to row \(3\).

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

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x+ 0 \cdot y = \frac{10}{19} \\ x = \frac{10}{19}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 1 \cdot y = \frac{9}{19} \\ y = \frac{9}{19}\end{aligned}$$$$\begin{aligned}0 \cdot x+ 0 \cdot y = 0 \\ None = 0\end{aligned}$$