Math Problem Solver

The system of equations has one solution.\[a = 2\] \[b = 3\] \[c = 5\] \[d = 2\]

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\1 & 0 & -1 & 0 &\bigm |& -3\\1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\1 & 0 & -1 & 0 &\bigm |& -3\\1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\1 & 0 & -1 & 0 &\bigm |& -3\\1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\0 & -1 & -1 & 0 &\bigm |& -8\\1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\0 & -1 & -1 & 0 &\bigm |& -8\\0 & -1 & 0 & -1 &\bigm |& -5\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\0 & -1 & -1 & 0 &\bigm |& -8\\0 & -1 & 0 & -1 &\bigm |& -5\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) to make the leading term \(1\).

$$\begin{bmatrix}1 & 1 & 0 & 0 &\bigm |& 5\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & -1 & 0 & -1 &\bigm |& -5\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & -1 & 0 &\bigm |& -3\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & -1 & 0 & -1 &\bigm |& -5\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

Multiply row \(2\) by scalar \(1\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & -1 & 0 &\bigm |& -3\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

Multiply row \(2\) by scalar \(-1\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & -1 & 0 &\bigm |& -3\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & -1 & 1 &\bigm |& -3\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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.

$$\begin{bmatrix}1 & 0 & -1 & 0 &\bigm |& -3\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & -1 & 1 &\bigm |& -3\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

The leading term of row \(3\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & 0 & -1 & 0 &\bigm |& -3\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & -1 & 1 &\bigm |& -3\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 1 & 0 &\bigm |& 8\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & -1 & 1 &\bigm |& -3\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

Multiply row \(3\) by scalar \(-1\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & -1 & 1 &\bigm |& -3\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

Multiply row \(3\) by scalar \(1\) and add it to row \(4\).

$$\begin{bmatrix}1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & 0 & 0 &\bigm |& 0\\0 & 0 & 1 & 1 &\bigm |& 7\end{bmatrix}$$

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

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

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.

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

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

$$\begin{bmatrix}1 & 0 & 0 & -1 &\bigm |& 0\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & 0 & 1 &\bigm |& 2\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

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

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& 2\\0 & 1 & 0 & 1 &\bigm |& 5\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & 0 & 1 &\bigm |& 2\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(4\) by scalar \(-1\) and add it to row \(2\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& 2\\0 & 1 & 0 & 0 &\bigm |& 3\\0 & 0 & 1 & -1 &\bigm |& 3\\0 & 0 & 0 & 1 &\bigm |& 2\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Multiply row \(4\) by scalar \(1\) and add it to row \(3\).

$$\begin{bmatrix}1 & 0 & 0 & 0 &\bigm |& 2\\0 & 1 & 0 & 0 &\bigm |& 3\\0 & 0 & 1 & 0 &\bigm |& 5\\0 & 0 & 0 & 1 &\bigm |& 2\\0 & 0 & 0 & 0 &\bigm |& 0\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot a+ 0 \cdot b+ 0 \cdot c+ 0 \cdot d = 2 \\ a = 2\end{aligned}$$$$\begin{aligned}0 \cdot a+ 1 \cdot b+ 0 \cdot c+ 0 \cdot d = 3 \\ b = 3\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b+ 1 \cdot c+ 0 \cdot d = 5 \\ c = 5\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b+ 0 \cdot c+ 1 \cdot d = 2 \\ d = 2\end{aligned}$$$$\begin{aligned}0 \cdot a+ 0 \cdot b+ 0 \cdot c+ 0 \cdot d = 0 \\ None = 0\end{aligned}$$