Lesson 11.6: Solving Systems with Gaussian Elimination

New Concept

Gaussian elimination is a powerful method for solving systems of linear equations. It involves representing the system as an augmented matrix and using row operations to simplify it into row-echelon form, which makes finding the solution straightforward.

What’s next

Next, you’ll practice converting systems to augmented matrices and applying row operations on interactive cards, preparing you to solve full systems step-by-step.

Property

To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix. A matrix containing just the coefficients is called the coefficient matrix. It is very important that each equation is written in standard form $ax + by + cz = d$ so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.

To write an augmented matrix:

1. Write the coefficients of the $x$-terms as the numbers down the first column.
2. Write the coefficients of the $y$-terms as the numbers down the second column.
3. If there are $z$-terms, write their coefficients down the third column.
4. Draw a vertical line and write the constants to the right.

Examples

• The system of equations $2x + 5y = 11$ and $3x - y = 8$ can be written as the augmented matrix $\begin{bmatrix} 2 & 5 & | & 11 \\ 3 & -1 & | & 8 \end{bmatrix}$.

Property

We can use the information in an augmented matrix to write the system of equations in standard form. Each column on the left of the vertical line represents the coefficients for a specific variable ($x, y, z, ...$), and the column on the right represents the constants. Each row corresponds to one equation in the system.

Examples

• The augmented matrix $\begin{bmatrix} 2 & -4 & | & 6 \\ 1 & 5 & | & -3 \end{bmatrix}$ represents the system of equations $2x - 4y = 6$ and $x + 5y = -3$.

• Given the matrix $\begin{bmatrix} 1 & 0 & 2 & | & 9 \\ 0 & 1 & -3 & | & 4 \\ 4 & 5 & 0 & | & 1 \end{bmatrix}$, the corresponding system is $x + 2z = 9$, $y - 3z = 4$, and $4x + 5y = 1$.

Property

To solve a system of equations, we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-substitution to find the solution.

• Interchange rows. (Notation: $R_i \leftrightarrow R_j$)
• Multiply a row by a constant. (Notation: $cR_i$)
• Add the product of a row multiplied by a constant to another row. (Notation: $R_j + cR_i$)

Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, we use Gaussian elimination to write a matrix in row-echelon form, which has ones on the main diagonal and zeros below them.

Property

We can solve a system of linear equations using matrices. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables. This method is known as Gaussian elimination.

To solve a system using matrices:

1. Write the system of equations as an augmented matrix.
2. Use row operations to convert the matrix to row-echelon form, where there are 1s on the main diagonal and 0s below it.
3. Write the system of equations corresponding to the row-echelon form.
4. Use back-substitution to find the solution.

Examples

• Solving the system $x+y=7$ and $2x-y=2$ starts with the matrix $\begin{bmatrix} 1 & 1 & | & 7 \\ 2 & -1 & | & 2 \end{bmatrix}$. After $-2R_1+R_2=R_2$, we get $\begin{bmatrix} 1 & 1 & | & 7 \\ 0 & -3 & | & -12 \end{bmatrix}$. This gives $-3y=-12$, so $y=4$. Back-substituting, $x+4=7$, so $x=3$. The solution is $(3, 4)$.