Math Problem Solver

The system of equations has one solution.$$B = -1$$ $$C = -5$$ $$D = 5$$ $$x = 5$$

Solve by Gauss-Jordan Elimination:

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

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.

$$\begin{bmatrix}5 & 0 & 0 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & -1 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

Multiply row $1$ by scalar $\frac{1}{5}$ to make the leading term $1$.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & -1 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & -1 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

The leading term of row $2$ is already $1$ so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & -1 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & -1 & 1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

Multiply row $3$ by scalar $-1$ to make the leading term $1$.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & 1 & -1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & 1 & -1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

The leading term of row $4$ is already $1$ so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 & 0 & \frac{1}{5} &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & 1 & -1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& -1\\0 & 1 & 0 & 1 &\bigm |& 0\\0 & 0 & 1 & -1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

Multiply row $4$ by scalar $-1$ and add it to row $2$.

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& -1\\0 & 1 & 0 & 0 &\bigm |& -5\\0 & 0 & 1 & -1 &\bigm |& 0\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

Multiply row $4$ by scalar $1$ and add it to row $3$.

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& -1\\0 & 1 & 0 & 0 &\bigm |& -5\\0 & 0 & 1 & 0 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\end{bmatrix}$$

Multiply row $4$ by scalar $-1$ and add it to row $5$.

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& -1\\0 & 1 & 0 & 0 &\bigm |& -5\\0 & 0 & 1 & 0 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 5\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot B+ 0 \cdot C+ 0 \cdot D+ 0 \cdot x = -1 \\ B = -1\end{aligned}$$$$\begin{aligned}0 \cdot B+ 1 \cdot C+ 0 \cdot D+ 0 \cdot x = -5 \\ C = -5\end{aligned}$$$$\begin{aligned}0 \cdot B+ 0 \cdot C+ 1 \cdot D+ 0 \cdot x = 5 \\ D = 5\end{aligned}$$$$\begin{aligned}0 \cdot B+ 0 \cdot C+ 0 \cdot D+ 1 \cdot x = 5 \\ x = 5\end{aligned}$$$$\begin{aligned}0 \cdot B+ 0 \cdot C+ 0 \cdot D+ 0 \cdot x = 0 \\ None = 0\end{aligned}$$