Lesson 4.5: Solve Systems of Equations Using Matrices

New Concept

Solving systems of equations can be simplified using a matrix, a rectangular array of numbers. This lesson teaches you to convert equations into an augmented matrix, perform row operations, and efficiently find the system's solution.

What’s next

You're all set! Next, you’ll tackle interactive examples on how to build and manipulate matrices to solve systems of equations step-by-step.

Property

A matrix is a rectangular array of numbers arranged in rows and columns.
To represent a system of linear equations with a matrix, we write each equation in standard form. The coefficients of the variables and the constant of each equation become a row in the matrix. A vertical line replaces the equal signs. This is the augmented matrix for the system.
For the system:

The augmented matrix is:

Examples

• The system $\begin{cases} 2x - 7y = 4 \\ x = 3y + 1 \end{cases}$ is first rewritten with the second equation in standard form as $x - 3y = 1$. The augmented matrix is $\left[\begin{array}{cc|c} 2 & -7 & 4 \\ 1 & -3 & 1 \end{array}\right]$.

• The system $\begin{cases} x + 2y - z = 5 \\ 3x + y + 4z = 8 \\ 2x - y + 9z = -2 \end{cases}$ corresponds to the augmented matrix $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 5 \\ 3 & 1 & 4 & 8 \\ 2 & -1 & 9 & -2 \end{array}\right]$.

Property

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

1. Interchange any two rows. ($R_i \leftrightarrow R_j$)
2. Multiply a row by any real number except 0. ($kR_i$)
3. Add a nonzero multiple of one row to another row. ($R_i + kR_j$)

Examples

Given the matrix $\left[\begin{array}{cc|c} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array}\right]$:

• Interchanging rows 1 and 2 ($R_1 \leftrightarrow R_2$) results in $\left[\begin{array}{cc|c} 2 & 4 & 6 \\ 1 & 3 & 5 \end{array}\right]$.

• Multiplying row 2 by $3$ ($3R_2$) results in $\left[\begin{array}{cc|c} 1 & 3 & 5 \\ 6 & 12 & 18 \end{array}\right]$.

Property

For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a $1$ and all entries below the diagonal are zeros.

Examples

• The matrix $\left[\begin{array}{cc|c} 1 & 5 & 2 \\ 0 & 1 & -3 \end{array}\right]$ is in row-echelon form. The diagonal entries are $1$ and the entry below the diagonal is $0$.

• The matrix $\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 5 & 6 \\ 0 & 7 & 1 & 8 \end{array}\right]$ is not in row-echelon form because the entry in row 3, column 2 is $7$, not $0$.

Property

To solve a system of equations using matrices:

1. Write the augmented matrix for the system of equations.
2. Using row operations, get the entry in row 1, column 1 to be 1.
3. Using row operations, get zeros in column 1 below the 1.
4. Using row operations, get the entry in row 2, column 2 to be 1.
5. Continue the process until the matrix is in row-echelon form.
6. Write the corresponding system of equations.
7. Use substitution to find the remaining variables.
8. Write the solution as an ordered pair or triple.
9. Check that the solution makes the original equations true.

Examples

• To solve $\begin{cases} x + 3y = 7 \\ 2x + 5y = 12 \end{cases}$, we form $\left[\begin{array}{cc|c} 1 & 3 & 7 \\ 2 & 5 & 12 \end{array}\right]$. The operation $-2R_1 + R_2$ gives $\left[\begin{array}{cc|c} 1 & 3 & 7 \\ 0 & -1 & -2 \end{array}\right]$. Multiplying $R_2$ by $-1$ gives the row-echelon form, which corresponds to $y=2$ and $x+3y=7$. Substituting gives $x=1$. The solution is $(1, 2)$.

• For the system $\begin{cases} 2x - 4y = 10 \\ x + y = 4 \end{cases}$, start by swapping rows to get a $1$ in the top-left corner: $R_1 \leftrightarrow R_2$ gives $\left[\begin{array}{cc|c} 1 & 1 & 4 \\ 2 & -4 & 10 \end{array}\right]$.

Property

When solving a system using matrices, if you obtain a row where all entries to the left of the vertical line are zero and the entry to the right is a non-zero number, the system is inconsistent. This row represents a false statement, such as $0 = 1$.

Examples

• The matrix $\left[\begin{array}{cc|c} 1 & 2 & 4 \\ 0 & 0 & 3 \end{array}\right]$ represents an inconsistent system. The second row implies $0 = 3$, which is false. There is no solution.

• The system $\begin{cases} x - 2y = 5 \\ -2x + 4y = -8 \end{cases}$ becomes $\left[\begin{array}{cc|c} 1 & -2 & 5 \\ -2 & 4 & -8 \end{array}\right]$. The operation $2R_1 + R_2$ yields $\left[\begin{array}{cc|c} 1 & -2 & 5 \\ 0 & 0 & 2 \end{array}\right]$, showing there is no solution.

Property

When solving a system using matrices, if you obtain a row where all entries are zero (both to the left and right of the vertical line), the system is dependent. This row represents a true but unhelpful statement, $0 = 0$, indicating there are infinitely many solutions.

Examples

• The matrix $\left[\begin{array}{cc|c} 1 & 3 & 5 \\ 0 & 0 & 0 \end{array}\right]$ represents a dependent system. The second row is $0=0$. The solutions can be described by the first row: $x+3y=5$.

• The system $\begin{cases} x + y = 4 \\ 3x + 3y = 12 \end{cases}$ becomes $\left[\begin{array}{cc|c} 1 & 1 & 4 \\ 3 & 3 & 12 \end{array}\right]$. The operation $-3R_1 + R_2$ yields $\left[\begin{array}{cc|c} 1 & 1 & 4 \\ 0 & 0 & 0 \end{array}\right]$, indicating infinite solutions.