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Solving Systems of Linear Equations Using Matrices
What is a Matrix?
A matrix is a compact grid or array of numbers. It can be created from a system of equations and
used to solve the system of equations. Matrices have many applications in science, engineering, and
math courses. This handout will focus on how to solve a system of linear equations using matrices.
How to Solve a System of Equations Using Matrices
Matrices are useful for solving systems of equations. There are two main methods of solving
systems of equations: Gaussian elimination and Gauss-Jordan elimination. Both processes begin the
same way. To begin solving a system of equations with either method, the equations are first
changed into a matrix. The coefficient matrix is a matrix comprised of the coefficients of the variables
which is written such that each row represents one equation and each column contains the
coefficients for the same variable in each equation. The constant matrix is the solution to each of the
equations written in a single column and in the same order as the rows of the coefficient matrix.
The augmented matrix is the coefficient matrix with the constant matrix as the last column.
Example : Write the coefficient matrix, constant matrix, and augmented matrix for the following system of equations: − 3 𝑥𝑥 − 2 𝑦𝑦 + 4𝑧𝑧 = 9 3 𝑦𝑦 − 2 𝑧𝑧 = 5 4 𝑥𝑥 − 3 𝑦𝑦 + 2𝑧𝑧 = 7 Solution : The coefficient matrix is created by taking the coefficients of each variable and entering them into each row. The first equation will be the first row; the second equation will be the second row, and the third equation will be the third row. Also, the first column will represent the “𝑥𝑥” variable; the second column will represent the “𝑦𝑦” variable, and the
third column will represent the “𝑧𝑧” variable.
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Because the second equation does not contain an “𝑥𝑥” variable, a “ 0 ” has been entered into the “𝑥𝑥” column in the second row.
The constant matrix is a single column matrix consisting of the solutions to the equations.
To create the augmented matrix, add the constant matrix as the last column of the coefficient matrix.
For the Gaussian elimination method, once the augmented matrix has been created, use elementary
row operations to reduce the matrix to Row-Echelon form. There are three basic types of
elementary row operations: (1) row swapping, (2) row multiplication, and (3) row addition. Row
multiplication and row addition can be combined together.
(1) In row swapping , the rows exchange positions within the matrix. The matrix resulting from a row
operation or sequence of row operations is called row equivalent to the original matrix.
Example : Swap row one and row three Solution :
𝑅𝑅 1 ↔𝑅𝑅 3 �⎯⎯⎯� �
(2) In row multiplication , every entry in a row is multiplied by the same constant.
Example : Multiply row one by −^13
Solution :
− 13 𝑅𝑅 1 �⎯⎯� �
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Solution b): Yes, this matrix is in Row-Echelon form as the leading entry in each row has 0’s below, and the leading entry in each row is to the right of the leading entry in the row above. Notice the leading entry for row three is in column 4 not column 3. The leading
entry is allowed to skip columns, but it cannot be to the left of the leading entry in any row above it.
Solution c): Yes, this matrix is in Row-Echelon form. Each leading entry in each row is to the right of the leading entry in the row above it, and each leading entry contains only 0’s below it.
The following example will demonstrate how to use the elementary row operations to reduce the
augmented matrix from a system of equations to Row-Echelon form. After Row-Echelon form is
achieved, back substitution can be used to find the solution to the system of equations.
Example : Solve the following system of equations using Gaussian Elimination: − 3 𝑥𝑥 − 2 𝑦𝑦 + 4𝑧𝑧 = 9 3 𝑦𝑦 − 2 𝑧𝑧 = 5 4 𝑥𝑥 − 3 𝑦𝑦 + 2𝑧𝑧 = 7 Solution : First, create the augmented matrix for the system.
Next use the elementary row operations to reduce the matrix to Row-Echelon form.
−^13 𝑅𝑅 1 �⎯� �
1 23 −^43 − 3
−4𝑅𝑅 1 +𝑅𝑅 3 �⎯⎯⎯⎯⎯� �
1 23 −^43 − 3
0 −^173223
− 173 𝑅𝑅 3 �⎯⎯�
1 23 −^43 − 3
0 1 −^2217 −^5717
𝑅𝑅 3 ↔𝑅𝑅 2 �⎯⎯⎯� �
1 23 −^43 − 3
0 1 −^2217 −^5717
−3𝑅𝑅 2 +𝑅𝑅 3 �⎯⎯⎯⎯⎯� �
1 23 −^43 − 3
0 1 −^2217 −^5717
17 32 𝑅𝑅^3 �⎯�
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1 23 −^43 − 3
0 1 −^2217 −^5717
Finally, rewrite the matrix as a system of reduced equations and back substitute to find the solution. 1 𝑥𝑥 + 2 3 𝑦𝑦 −^
4 3 𝑧𝑧^ =^ −^3 1 𝑦𝑦 − 2217 𝑧𝑧 = −^5717 1 𝑧𝑧 = 8
The reduced equations show that 𝑧𝑧 = 8. Substitute 8 for 𝑧𝑧 and solve for 𝑦𝑦 in the second equation.
1 𝑦𝑦 − 22 17 (8) =^ −
57 17 𝑦𝑦 − 17617 = −^5717
𝑦𝑦 = 7
Substitute 8 for 𝑧𝑧 and 7 for 𝑦𝑦 in the first equation and solve for 𝑥𝑥.
𝑥𝑥 + 23 (7) − 43 (8) = − 3
𝑥𝑥 + 14 3 −^
32 3 =^ −^3 𝑥𝑥 − 6 = − 3 𝑥𝑥 = 3 The solution to the system of equations is (3,7,8).
An alternative method, the Gauss-Jordan elimination method, can be used to solve the system of
equations. This involves reducing the augmented matrix to Reduced Row-Echelon form. The
Reduced Row-Echelon form is similar to the Row-Echelon form except that the leading entry in
each row must be a 1 and all other entries in the same column as a leading entry must be 0. Unlike
the Row-Echelon form, there is one and only one Reduced Row-Echelon form for a system of
equations.
Example : Solve the following system of equations using Gauss-Jordan Elimination:
− 3 𝑥𝑥 − 2 𝑦𝑦 + 4𝑧𝑧 = 9
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when it is row reduced to either Row-Echelon form or Reduced Row-Echelon form. In other
words, each variable represented by a column can be solved for a specific number. With an
inconsistent system of equations , the leading coefficient in one of the rows will be in the last column of
the augmented matrix.
Example : Determine if the following system of equations is consistent or inconsistent and state the solution. 2 𝑥𝑥 − 4 𝑦𝑦 + 𝑧𝑧 = 3 𝑥𝑥 − 3 𝑦𝑦 + 𝑧𝑧 = 5 3 𝑥𝑥 − 7 𝑦𝑦 + 2𝑧𝑧 = 12
Solution : First, create the augmented matrix.
Use the elementary row operations to obtain a Row-Echelon form.
𝑅𝑅 2 ↔𝑅𝑅 1 �⎯⎯⎯� �
−2𝑅𝑅 1 +𝑅𝑅 2 �⎯⎯⎯⎯⎯� �
−3𝑅𝑅 1 +𝑅𝑅 3 �⎯⎯⎯⎯⎯�
−1𝑅𝑅 2 +𝑅𝑅 3 �⎯⎯⎯⎯⎯� �
1 (^2) �𝑅𝑅2� �
0 1 −^12 −^72
The last row indicates the system is inconsistent. This can most easily be seen if the last row is converted back to an equation. 0 𝑥𝑥 + 0𝑦𝑦 + 0𝑧𝑧 = 4 According to this equation, there are not any values of 𝑥𝑥, 𝑦𝑦, or 𝑧𝑧 that will make the above equation true. Therefore, the system has no solution; this is represented by the symbol for the null set, ∅. Any augmented system of equations is inconsistent if the Row-Echelon form
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contains a row with the coefficient portion of the row containing all 0’s and the augmented column containing any number except 0.
A system of equations can also be dependent. In the case of a dependent system , one of the columns of
the coefficient portion of the augmented matrix will lack a leading coefficient. In some cases, the
row corresponding to the missing leading coefficient will contain only 0’s. In other cases, there will
be fewer rows than columns. Be careful because the presence of a row of 0’s does not automatically
indicate a dependent system. If a system of three equations contains only two variables, then a row
of 0’s does not indicate a dependent system. However, a system of three equations with three
variables that contains a row of 0s indicates a dependent system.
Example 1 : Determine if the following system of equations is consistent, inconsistent, or dependent, and state the solution. 2 𝑥𝑥 − 3 𝑦𝑦 = − 21 3 𝑥𝑥 + 2𝑦𝑦 = 1 8 𝑥𝑥 − 5 𝑦𝑦 = − 49 Solution : Create the augmented matrix and use elementary row operations to reduce the ` matrix to Row-Echelon form.
1 (^2) �𝑅𝑅1� �
1 −^32 −^212
−3𝑅𝑅1+𝑅𝑅 �⎯⎯⎯⎯� �
1 −^32 −^212
−8𝑅𝑅 1 +𝑅𝑅 3 �⎯⎯⎯⎯⎯�
1 −^32 −^212
2 (^13) �⎯𝑅𝑅�^2 �
1 −^32 −^212
−7𝑅𝑅 2 +𝑅𝑅 3 �⎯⎯⎯⎯⎯� �
1 −^32 −^212
3 (^2) �𝑅𝑅⎯^2 ⎯+𝑅𝑅⎯�^1 �
This system of equations is consistent even though there is a row of 0 ’s at the bottom. The original system of equations had only two variables, 𝑥𝑥 and 𝑦𝑦, thus the solution to the system
only contains two numbers. The solution to the system of equations is (−3,5).
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The solution to the dependent system of equations is (− 2 𝑧𝑧 + 5, 𝑧𝑧 − 2, 𝑧𝑧).
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Practice Problems
Solve the following systems of equations by:
Gaussian Elimination Gauss-Jordan Elimination:
- 2 𝑥𝑥 − 3 𝑦𝑦 + 2𝑧𝑧 = 13 4. 𝑥𝑥 − 3 𝑦𝑦 + 𝑧𝑧 = 8 3 𝑥𝑥 + 𝑦𝑦 − 𝑧𝑧 = 2 2 𝑥𝑥 − 5 𝑦𝑦 − 3 𝑧𝑧 = 2 3 𝑥𝑥 − 4 𝑦𝑦 − 3 𝑧𝑧 = 1 𝑥𝑥 + 4𝑦𝑦 + 𝑧𝑧 = 2
Solutions
- (2, −1,3)
- No Solution
- ( 27 2 𝑐𝑐^ + 39,^
5 2 𝑐𝑐^ + 10,^ −^4 𝑐𝑐 −^ 10,^ 𝑐𝑐)
- � 12 5 ,^ −1,^
13 5 �
- (16𝑐𝑐, 6𝑐𝑐, 𝑐𝑐)
- No Solution
