Math Problem Solver

The system of equations has one solution.\[x = 7\]

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}1 &\bigm |& 7\\1 &\bigm |& 7\end{bmatrix}$$

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.

$$\begin{bmatrix}1 &\bigm |& 7\\1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 &\bigm |& 7\\1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 &\bigm |& 7\\0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x = 7 \\ x = 7\end{aligned}$$$$\begin{aligned}0 \cdot x = 0 \\ None = 0\end{aligned}$$