8.3 Solving Linear Systems Using Matrices

In this section, we consider a mathematical tool called a matrix (plural: matrices) that has wide application in mathematics, business, science, and engineering. A matrix is a rectangular array of numbers or entries. These entries are ordinarily displayed in rows and columns, and the entire matrix is enclosed in brackets or parentheses. Thus,

$[ 1, 2, 3; 4, 5, 6; 7, 8, 9 ], [ 2, minus 1, 3; 4, 0, 2 ], and [ 4; 5; 6 ]$

are matrices. A matrix of order, or dimension, $n \times m$ (read " $n$ by $m$ ") has $n$ (horizontal) rows and $m$ (vertical) columns.

The matrices above are $3 times 3$ , $2 times 3$ , and $3 times 1$ , respectively. The first matrix—in which the number of rows is equal to the number of columns—is an example of a square matrix.

Coefficient Matrix and Augmented Matrix of a System

We will use matrices to solve systems of linear equations. For a system of linear equations of the form

$\begin{aligned} a_{1} x + b_{1} y + c_{1} z & = d_{1} \\ a_{2} x + b_{2} y + c_{2} z & = d_{2} \\ a_{3} x + b_{3} y + c_{3} z & = d_{3} \end{aligned}$

the matrices

$[ a sub 1, b sub 1, c sub 1; a sub 2, b sub 2, c sub 2; a sub 3, b sub 3, c sub 3 ] and [ a sub 1, b sub 1, c sub 1, d sub 1; a sub 2, b sub 2, c sub 2, d sub 2; a sub 3, b sub 3, c sub 3, d sub 3 ]$

are called the coefficient matrix and the augmented matrix, respectively. Each row of the augmented matrix represents one of the equations of the system. For example, the augmented matrix of the system

$\begin{aligned} 3 x & - & 4 y & + & z & = & 2 \\ - x & + & 2 y & = & - 1 \\ 2 x & - & y & - & 3 z & = & 4 \end{aligned}$

$[\begin{array}{cccc} 3 & - 4 & 1 & 2 \\ - 1 & 2 & 0 & - 1 \\ 2 & - 1 & 3 & 4 \end{array}]$

and the augmented matrix of the system

$\begin{aligned} x & - & 3 y & + & 2 z & = & 5 \\ 2 y & - & z & = & 4 \\ 4 z & = & 8 \end{aligned}$

$[\begin{array}{cccc} 1 & - 3 & 2 & 5 \\ 0 & 2 & - 1 & 4 \\ 0 & 0 & 4 & 8 \end{array}]$

The augmented matrix of this last system, which has all zero entries in the lower left corner (below the diagonal), is said to be in upper triangular form. As we saw in Systems of Linear Equations in Three Variables, it is easy to find the solution of such a system by back-substitution.

What is the difference between a coefficient matrix and an augmented matrix?

_____

What is the difference between a coefficient matrix and an augmented matrix?

Elementary Row Operations

The method of elimination depends on two properties of linear systems that allow us to change one system into an equivalent one, that is, one that has the same solutions as the original system.

Note

Properties of Linear Systems

Multiplying a linear equation by a (nonzero) constant does not change its solutions. That is, any solution of the equation
$a x + b y = c$
is also a solution of the equation
$k a x + k b y = k c$
Adding (or subtracting) two linear equations does not change their common solutions. That is, any solution of the system
$\begin{aligned} a_{1} x + b_{1} y & = c_{1} \\ a_{2} x + b_{2} y & = c_{2} \end{aligned}$
is also a solution of the equation
$(a_{1} + a_{2}) x + (b_{1} + b_{2}) y = c_{1} + c_{2}$

These properties apply to linear systems of any size. Thus, we can perform the following operations on the equations of a system:

Multiply both sides of an equation by a nonzero real number.
Add a constant multiple of one equation to another equation.
Interchange two equations.

Because each equation of the system corresponds to a row in the augmented matrix, the three operations above correspond to three elementary row operations for the augmented matrix. We can perform one or more of these operations on a matrix without changing the solution of the system it represents.

Which of the following is not an allowed row operation?

_____

Interchange two elements in a row

Which of the following is not an allowed row operation?

Divide the entries in a row by a nonzero number
Add two rows together
Interchange two elements in a row
Interchange the first and last rows

Example 1

$A = [\begin{array}{cccc} 1 & 3 & - 1 & - 1 \\ 2 & 1 & 4 & 5 \\ 6 & 2 & - 1 & - 12 \end{array}]$ and $B = [\begin{array}{cccc} 1 & 3 & - 1 & - 1 \\ 6 & 3 & 12 & 15 \\ 6 & 2 & - 1 & - 12 \end{array}]$
represent equivalent systems because we can multiply each entry in the second row of $A$ by $3$ to obtain $B$ .
$A = [\begin{array}{cccc} 3 & - 1 & 2 & 7 \\ 2 & 1 & 4 & - 5 \\ 3 & 1 & 9 & - 16 \end{array}]$ and $B = [\begin{array}{cccc} 3 & 1 & 9 & - 16 \\ 2 & 1 & 4 & - 5 \\ 3 & - 1 & 2 & 7 \end{array}]$
represent equivalent systems because we can interchange the first and third rows of $A$ to obtain $B$ .
$A = [\begin{array}{cccc} 1 & 2 & 1 & - 3 \\ 2 & 0 & - 1 & 7 \\ 3 & 1 & 2 & 10 \end{array}]$ and $B = [\begin{array}{cccc} 1 & 2 & 1 & - 3 \\ 0 & - 4 & - 3 & 13 \\ 3 & 1 & 2 & 10 \end{array}]$
represent equivalent systems becausewe can add $minus 2$ times each entry of the first row of $A$ to the corresponding entry of the second row of $A$ to obtain $B$ .

Subtract $4$ times row 2 from row 3.

$[\begin{array}{cccc} 3 & 2 & - 5 & - 3 \\ 2 & - 3 & 2 & 6 \\ 8 & - 4 & 2 & 12 \end{array}]$

The new row 3 becomes

[

_____

|

_____

]

$[\begin{array}{cccc} 3 & 2 & - 5 & - 3 \\ 2 & - 3 & 2 & 6 \\ 0 & 8 & - 6 & - 12 \end{array}]$

Subtract $4$ times row 2 from row 3.

$[\begin{array}{cccc} 3 & 2 & - 5 & - 3 \\ 2 & - 3 & 2 & 6 \\ 8 & - 4 & 2 & 12 \end{array}]$

$[\begin{array}{cccc} 3 & 2 & - 5 & - 3 \\ 2 & - 3 & 2 & 6 \\ 0 & 8 & - 6 & - 12 \end{array}]$

Example 2

Use row operations to form an equivalent matrix with the given elements:

$[\begin{array}{ccc} 1 & - 4 & - 5 \\ 3 & 6 & 3 \end{array}] ⟶ [\begin{array}{ccc} 1 & - 4 & - 5 \\ 0 & ? & ? \end{array}]$

To obtain $0$ as the first entry in the second row, we can multiply the first row by $minus 3$ and add the result to the second row. That is, we add $- 3 (row 1)$ to $row 2$ :

$[\begin{array}{ccc} 1 & - 4 & - 5 \\ 3 & 6 & 3 \end{array}] \to_{- 3 (row 1) + row 2}^{} [\begin{array}{ccc} 1 & - 4 & - 5 \\ 0 & 18 & 18 \end{array}]$

Use row operations to form an equivalent matrix with the given elements:

$[\begin{array}{ccc} 1 & - 3 & 7 \\ - 2 & 4 & - 6 \end{array}] ⟶ [\begin{array}{ccc} 1 & - 3 & 7 \\ 0 & ? & ? \end{array}]$

We add _____ times row 1 to row 2.

The new row 2 becomes

[ 0

_____

|

_____

]

$2 (row 1) + row 2 : [\begin{array}{ccc} 1 & - 3 & 7 \\ 0 & - 2 & 8 \end{array}]$

Use row operations to form an equivalent matrix with the given elements:

$[\begin{array}{ccc} 1 & - 3 & 7 \\ - 2 & 4 & - 6 \end{array}] ⟶ [\begin{array}{ccc} 1 & - 3 & 7 \\ 0 & ? & ? \end{array}]$

$2 (row 1) + row 2 : [\begin{array}{ccc} 1 & - 3 & 7 \\ 0 & - 2 & 8 \end{array}]$

Example 3

Use row operations on the first matrix to form an equivalent matrix in upper triangular form.

$[\begin{array}{cccc} 1 & - 3 & 1 & - 4 \\ 3 & - 1 & - 1 & 8 \\ 2 & - 2 & 3 & - 1 \end{array}] ⟶ [\begin{array}{cccc} 1 & - 3 & 1 & - 4 \\ 0 & ? & ? & ? \\ 0 & 0 & ? & ? \end{array}]$

We perform the transformation in two steps. First, we obtain zeros in the lower two entries of the first column. We get these zeros by adding suitable multiples of the first row to the second and third rows:

$[\begin{array}{cccc} 1 & - 3 & 1 & - 4 \\ 3 & - 1 & - 1 & 8 \\ 2 & - 2 & 3 & - 1 \end{array}] \to_{- 2 (row 1) + row 3}^{- 3 (row 1) + row 2} [\begin{array}{cccc} 1 & - 3 & 1 & - 4 \\ 0 & 8 & - 4 & 20 \\ 0 & 4 & 1 & 7 \end{array}]$

Next we obtain a zero as the second entry of the third row by adding a suitable multiple of the second row. For this matrix, we add $- 12 (row 2)$ to $row 3$ :

$[\begin{array}{cccc} 1 & - 3 & 1 & - 4 \\ 0 & 8 & - 4 & 20 \\ 0 & 4 & 1 & 7 \end{array}] \to_{\frac{- 1}{2} (row 2) + row 3}^{} [\begin{array}{cccc} 1 & - 3 & 1 & - 4 \\ 0 & 8 & - 4 & 20 \\ 0 & 0 & 3 & - 3 \end{array}]$

The last matrix is in upper triangular form.

Use the suggested row operations to form an equivalent matrix in upper triangular form.

$[\begin{array}{cccc} 1 & - 2 & 4 & 3 \\ 5 & - 7 & 8 & 6 \\ - 2 & 6 & - 7 & 6 \end{array}] ⟶ [\begin{array}{cccc} 1 & - 2 & 4 & 3 \\ 0 & ? & ? & ? \\ 0 & 0 & ? & ? \end{array}]$

Add $- 5 (row 1)$ to $open parenthesis row 2 close parenthesis$ .
The new row 2 becomes
$[ 0$ _____ _____ $|$ _____ $]$
Add $2 open parenthesis row 1 close parenthesis$ to $open parenthesis row 3 close parenthesis$ .
The new row 3 becomes
$[ 0$ _____ _____ $|$ _____ $]$
Multiply $open parenthesis row 2 close parenthesis$ by $\frac{1}{3}$ .
The new row 2 becomes
$[ 0$ _____ _____ $|$ _____ $]$
Add $- 2 (row 2)$ to $open parenthesis row 3 close parenthesis$ .
The new row 3 becomes
$[ 0$ $0$ _____ $|$ _____ $]$

$[\begin{array}{cccc} 1 & - 2 & 4 & 3 \\ 0 & 1 & - 4 & - 3 \\ 0 & 0 & 9 & 18 \end{array}]$

Use the suggested row operations to form an equivalent matrix in upper triangular form.

$[\begin{array}{cccc} 1 & - 2 & 4 & 3 \\ 5 & - 7 & 8 & 6 \\ - 2 & 6 & - 7 & 6 \end{array}] ⟶ [\begin{array}{cccc} 1 & - 2 & 4 & 3 \\ 0 & ? & ? & ? \\ 0 & 0 & ? & ? \end{array}]$

Add $- 5 (row 1)$ to $open parenthesis row 2 close parenthesis$ .
Add $2 open parenthesis row 1 close parenthesis$ to $open parenthesis row 3 close parenthesis$ .
Multiply $open parenthesis row 2 close parenthesis$ by $\frac{1}{3}$ .
Add $- 2 (row 2)$ to $open parenthesis row 3 close parenthesis$ .

$[\begin{array}{cccc} 1 & - 2 & 4 & 3 \\ 0 & 1 & - 4 & - 3 \\ 0 & 0 & 9 & 18 \end{array}]$

Matrix Reduction

We use elementary row operations to apply Gaussian reduction to the augmented matrix of a linear system. This matrix method can be used to solve linear systems of any size and is well suited for implementation by a computer program. The method has three steps.

Example 4

Use matrix reduction to solve the system.

$\begin{aligned} x & - & 2 y & = & - 5 \\ 2 x & + & 3 y & = & 11 \end{aligned}$

The augmented matrix is

$[\begin{array}{ccc} 1 & - 2 & - 5 \\ 2 & 3 & 11 \end{array}]$

We use row operations to obtain $0$ in the first entry of the second row.

$[\begin{array}{ccc} 1 & - 2 & - 5 \\ 2 & 3 & 11 \end{array}] \to_{- 2 (row 1) + row 2}^{} [\begin{array}{ccc} 1 & - 2 & - 5 \\ 0 & 7 & 21 \end{array}]$

The last matrix is upper triangular and corresponds to the system

$\begin{aligned} x & - & 2 y & = & - 5 & (1) \\ 7 y & = & 21 & (2) \end{aligned}$

Now we use back-substitution to solve the system. From Equation (2), $y equals 3$ . Substitute $3$ for $y$ in Equation (1) to find

$\begin{aligned} x - 2 (3) & = - 5 \\ x & = 1 \end{aligned}$

The solution is the ordered pair $open parenthesis 1, 3 close parenthesis$ .

Use matrix reduction to solve the system

$\begin{aligned} x & + & 4 y & = & 3 \\ 3 x & + & 8 y & = & 1 \end{aligned}$

Follow the suggested steps:

Write the augmented matrix for the system.
Add $- 3 (row 1)$ to $open parenthesis row 2 close parenthesis$ .
Solve the resulting system by back-substitution.

Solution: _____ Give the solution as an ordered pair.

$open parenthesis minus 5, 2 close parenthesis$

Use matrix reduction to solve the system

$\begin{aligned} x & + & 4 y & = & 3 \\ 3 x & + & 8 y & = & 1 \end{aligned}$

Follow the suggested steps:

Write the augmented matrix for the system.
Add $- 3 (row 1)$ to $open parenthesis row 2 close parenthesis$ .
Solve the resulting system by back-substitution.

$open parenthesis minus 5, 2 close parenthesis$

Why do we want to put a matrix into upper triangular form?

_____

Why do we want to put a matrix into upper triangular form?

Reducing a $3 times 3$ Matrix

Although there are many ways to reduce an augmented $3 times 3$ matrix to upper triangular form, the following two-step procedure may help you organize the row operations.

If you cannot obtain nonzero entries on the diagonal, then the system does not have a unique solution; the system is either inconsistent or dependent. See Homework Problems 45–48.

What happens if we cannot obtain nonzero entries on the diagonal?

_____

There is no unique solution.

What happens if we cannot obtain nonzero entries on the diagonal?

There is no solution.
There is no unique solution.
We should reduce the matrix.
We should augment the matrix.

Example 5

Use matrix reduction to solve the system.

$\begin{aligned} 2 x & - & 4 y & = & 6 \\ 3 x & - & 4 y & + & z & = & 8 \\ 2 x & - & 3 z & = & - 11 \end{aligned}$

The augmented matrix is

$[\begin{array}{cccc} 2 & - 4 & 0 & 6 \\ 3 & - 4 & 1 & 8 \\ 2 & 0 & - 3 & - 11 \end{array}]$

We obtain $1$ as the first entry of the first row by multiplying each entry in the first row by $\frac{1}{2}$ . This will make it easier to obtain zeros in the first entries of the second and third rows.

$[\begin{array}{cccc} 2 & - 4 & 0 & 6 \\ 3 & - 4 & 1 & 8 \\ 2 & 0 & - 3 & - 11 \end{array}]$

$\begin{aligned} 2 x & - & 4 y & + & 0 z & = & 6 \\ 3 x & - & 4 y & + & z & = & 8 \\ 2 x & + & 0 y & - & 3 z & = & - 11 \end{aligned}$

$\overset{\frac{1}{2} (row 1)}{↓}$

$[\begin{array}{cccc} 1 & - 2 & 0 & 3 \\ 3 & - 4 & 1 & 8 \\ 2 & 0 & - 3 & - 11 \end{array}]$

$\begin{aligned} x & - & 2 y & + & 0 z & = & 3 \\ 3 x & - & 4 y & + & z & = & 8 \\ 2 x & + & 0 y & - & 3 z & = & - 11 \end{aligned}$

Next, we obtain zeros in the first entries of the second and third rows by adding suitable multiples of the first row:

$- 3 (row 1) + 2 ↓ 3 (row 1) + 3$

$[\begin{array}{cccc} 1 & - 2 & 0 & 3 \\ 0 & 2 & 1 & - 1 \\ 0 & 4 & - 3 & - 17 \end{array}]$

$\begin{aligned} x & - & 2 y & + & 0 z & = & 3 \\ x & + & 2 y & + & z & = & - 1 \\ 0 x & + & 4 y & - & 3 z & = & - 17 \end{aligned}$

Finally, we obtain a zero in the second entry of the third row by adding $- 2 (row 2)$ to $row 3$ :

$- 2 (row 2) + 3 ↓$

$[\begin{array}{cccc} 1 & - 2 & 0 & 3 \\ 0 & 2 & 1 & - 1 \\ 0 & 0 & - 5 & - 15 \end{array}]$

$\begin{aligned} x & - & 2 y & + & 0 z & = & 3 \\ 0 x & + & 2 y & + & z & = & - 1 \\ 0 x & + & 0 y & - & 5 z & = & - 15 \end{aligned}$

The system is now in upper triangular form, and we use back-substitution to find the solution. We solve the last equation to get $z equals 3$ and substitute $3$ for $z$ in the second equation to find $y = - 2$ . Finally, we substitute $3$ for $z$ and $minus 2$ for $y$ in the first equation to find $x = - 1$ . The solution is the ordered triple $(- 1, - 2, 3)$ .

Use matrix reduction to solve the system

$\begin{aligned} x + 3 z & = - 11 \\ 2 x + y + z & = 1 \\ - 3 x - 2 y = 3 \end{aligned}$

Follow the suggested steps:

Write the augmented matrix for the system.
Add $minus 2$ (row 1) to (row 2).
Add $3$ (row 1) to (row 3).
Add $2$ (row 2) to (row 3).
Solve the resulting system by back-substitution.

Solution: _____ Give the solution as an ordered triple.

$(37, - 57, - 16)$

Use matrix reduction to solve the system

$\begin{aligned} x + 3 z & = - 11 \\ 2 x + y + z & = 1 \\ - 3 x - 2 y & = 3 \end{aligned}$

Follow the suggested steps:

Write the augmented matrix for the system.
Add $minus 2$ (row 1) to (row 2).
Add $3$ (row 1) to (row 3).
Add $2$ (row 2) to (row 3).
Solve the resulting system by back-substitution.

$(37, - 57, - 16)$

It is a good idea to make the first entry in the first row equal to $1$ , as we did in Example, to simplify the calculations that follow. However, if that entry is zero we can interchange two rows. For example, to reduce the matrix

$[ 0, 1, 2, 3; 1, 2, 3, 4; 2, 5, 1, 4 ]$

we begin by interchanging the first and second rows to obtain

$[ 1, 2, 3, 4; 0, 1, 2, 3; 2, 5, 1, 4 ]$

We then follow the two-step procedure described above to find

$\to_{- 2 (row 1) + row 3}^{} [\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & - 5 & - 4 \end{array}] \to_{- (row 2) + row 3}^{} [\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & - 7 & - 7 \end{array}]$

What should you do if the first entry in the first row is zero?

_____

Interchange two rows

What should you do if the first entry in the first row is zero?

Interchange two columns
Interchange two rows
Add the first row to the second row
Obtain zeros below that entry

Solving Larger Systems

Systems of linear equations in business and economics may involve hundreds of linear equations and hundreds of variables. The study of airflow for aircraft design can involve systems of millions of linear equations and variables. These large systems are solved using matrix methods.

Example 6

Write the augmented matrix for the system

$\begin{aligned} - w & + & x & - & y & + & z & = & 7 \\ 2 w & - & 2 x & - & 4 y & - & 5 z & = & - 17 \\ - 3 w & + & 9 x & + & 3 y & + & 6 z & = & 30 \\ 4 w & + & 8 x & - & 2 y & + & 5 z & = & 17 \end{aligned}$

Use matrix reduction and back-substitution to solve the system.

The augmented matrix is

$[\begin{array}{ccccc} - 1 & 1 & - 1 & 1 & 7 \\ 2 & - 2 & - 4 & - 5 & - 17 \\ - 3 & 9 & 3 & 6 & 30 \\ 4 & 8 & - 2 & 5 & 17 \end{array}]$

We use row operations to put the matrix in upper triangular form. First, we add multiples of the first row to the rows below it to get $0$ s in the first column.

2 (row 1) + row 2

- 3 (row 1) + row 3

4 (row 1) + row 4

$[\begin{array}{ccccc} - 1 & 1 & - 1 & 1 & 7 \\ 0 & 0 & - 6 & - 3 & - 3 \\ 0 & 6 & 6 & 3 & 9 \\ 0 & 12 & - 6 & 9 & 45 \end{array}]$

The second row has $0$ as its second entry, so we interchange the second and third rows to put a nonzero entry in the second column.

$[\begin{array}{ccccc} - 1 & 1 & - 1 & 1 & 7 \\ 0 & 6 & 6 & 3 & 9 \\ 0 & 0 & - 6 & - 3 & - 3 \\ 0 & 12 & - 6 & 9 & 45 \end{array}]$

Now we want $0$ s in the last two rows of the second column. The third row already has $0$ in the second column, so we only need to add a multiple of the second row to the fourth row.

- 2 (row 2) + row 4

$[\begin{array}{ccccc} - 1 & 1 & - 1 & 1 & 7 \\ 0 & 6 & 6 & 3 & 9 \\ 0 & 0 & - 6 & - 3 & - 3 \\ 0 & 0 & - 18 & 3 & 27 \end{array}]$

Finally, we add a multiple of the third row to the fourth row to get 0 in the third column.

- 3 (row 3) + row 4

$[\begin{array}{ccccc} - 1 & 1 & - 1 & 1 & 7 \\ 0 & 6 & 6 & 3 & 9 \\ 0 & 0 & - 6 & - 3 & - 3 \\ 0 & 0 & 0 & 12 & 36 \end{array}]$

This triangular matrix corresponds to the system

$\begin{aligned} - w & + & x & - & y & + & z & = & 7 \\ 6 x & + & 6 y & + & 3 z & = & 9 \\ 6 y & + & 3 z & = & - 3 \\ 12 z & = & 36 \end{aligned}$

which we solve by back-substitution.

From the last equation, we find $z equals 3$ , and by substituting $3$ for $z$ in the third equation, we find $y = - 1$ . Next, we substitute $minus 1$ for $y$ and $3$ for $z$ in the second equation to find $x equals 1$ , and finally, we substitute $x equals 1$ , $y = - 1$ , $z equals 3$ into the first equation to find $w = - 2$ .

The solution is the ordered four-tuple $(- 2, 1, - 1, 3)$ . We can verify that these values satisfy all four equations of the original system.

Our strategy for reducing $3 times 4$ augmented matrices can be generalized for larger matrices, as demonstrated in Example. Starting with the first row, we work our way along the diagonal, using row operations to obtain nonzero entries on the diagonal and zeros below the diagonal entry.

Use matrix reduction to solve the system

$\begin{aligned} w & + & x & + & y & + & z & = & - 1 \\ - w & + & x & - & y & + & z & = & 3 \\ - 2 w & + & 10 x & + & 4 y & + & 7 z & = & 23 \\ 3 w & + & 9 x & - & 3 y & + & 6 z & = & 0 \end{aligned}$

Solution: _____ Give the solution as an ordered 4-tuple.

$(- 3, 2, 1, - 1)$

Use matrix reduction to solve the system

$(- 3, 2, 1, - 1)$

How is matrix reduction similar to Gaussian reduction?

_____

How is matrix reduction similar to Gaussian reduction?

Reduced Row Echelon Form

The triangular matrix we obtain by row reduction is an example of row echelon form. In a row echelon matrix,

The first nonzero entry in any row is farther to the right than the first nonzero entry in the rows above it.
The entries directly below the first nonzero entry in any row are all zeros.
Any row consisting entirely of zeros is below all rows with any nonzero entry.

If we divide each row by its leading nonzero entry, so that the leading entry is $1$ , and if we use row reduction to obtain 0's above the leading $1$ as well as below it, the resulting matrix is said to be in reduced row echelon form. This form is especially convenient because we do not have to use back-substitution to finish solving the system; we can read off the solutions directly from the matrix.

For example, you can check that the matrix

$[\begin{array}{ccccc} 1 & 0 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & - 3 \\ 0 & 0 & 0 & 1 & 4 \end{array}]$

corresponds to the system

$\begin{aligned} a & + & 0 b & + & 0 c & + & 0 d & = & - 2 \\ 0 a & + & b & + & 0 c & + & 0 d & = & 3 \\ 0 a & + & 0 b & + & c & + & 0 d & = & - 3 \\ 0 a & + & 0 b & + & 0 c & + & d & = & 4 \end{aligned}$

From this system, it is easy to read the solutions $a = - 2$ , $b equals 3$ , $c = - 3$ , and $d equals 4$

Why is reduced row echelon form useful?

_____

It lets us read the solutions of a system.

Example 7

Solve the system

$\begin{aligned} a & + & 2 b & + & 4 c & + & 8 d & = & 12 \\ - 2 a & + & 2 b & - & 2 c & + & 2 d & = & 1 \\ 6 a & + & 6 b & + & 6 c & + & 6 d & = & 19 \\ 4 a & + & 20 b & - & 8 c & + & 14 d & = & 41 \end{aligned}$

by finding the reduced row echelon form of the augmented matrix.

The augmented matrix is

$[\begin{array}{ccccc} 1 & 2 & 4 & 8 & 12 \\ - 2 & 2 & - 2 & 2 & 1 \\ 6 & 6 & 6 & 6 & 19 \\ 4 & 20 & - 8 & 14 & 41 \end{array}]$

We enter this matrix into the calculator as follows: First access the MATRIX menu by pressing 2nd on a TI-84 orMATRX on a TI-83. You will see the menu shown in figure (a). We will use matrix [A], which is already selected, and we presss ENTER to EDIT (or enter) the matrix, shown in figure(b).

We want to enter a $4 times 5$ matrix, so we press $4$ ENTER $5$ ENTER, and we see the display in figure (a) below. Now type in the first row of the matrix, pressing ENTER after each entry. The calculator automatically moves to the second row. Continue filling in the rest of the augmented matrix, as shown in figure (b).

To make sure you have entered the values correctly, press 2ndMODE to quit to the home screen, then open the matrix menu again. Press $1$ ENTER to retrieve matrix [A]; the calculator display should look like figure (a) below. To check the rest of the matrix, press the right arrow key until you see the last column, as in figure (b).

Now we are ready to compute the reduced row echelon form of the matrix. Access the matrix menu again, but this time press the right arrow once to highlight MATH as shown in figure (a). Scroll down until the rref( command is highlighted, as shown in figure (b), and press ENTER.

Finally, enter matrix [A] after the rref( command by pressing 2nd $1$ ) ENTER or $M A T R X$ $1$ ) ENTER. The display should now look like figure (a) below. The last column of the matrix gives decimal approximations for the solutions.

The calculator can also display the rational form of the solutions: Press MATH ENTER ENTER to see figure(b). The reduced row echelon form of the augmented matrix is thus

$[ 1, 0, 0, 0, 2; 0, 1, 0, 0, 1 / 3; 0, 0, 1, 0, minus 2 / 3; 0, 0, 0, 1, 3 / 2 ]$

and the solution is the ordered four-tuple $(2, \frac{1}{3}, \frac{- 2}{3}, \frac{3}{2})$ . You can verify that these values satisfy each of the original four equations.

Find the row reduced echelon form of the augmented matrix for the system
$\begin{aligned} - a & + & b & - & c & + & d & = & 8 \\ a & + & b & + & c & + & d & = & 2 \\ 8 a & + & 4 b & + & 2 c & + & d & = & 5 \\ 27 a & + & 9 b & - & 3 c & + & d & = & 4 \end{aligned}$
What third degree polynomial has a graph that passes through the points $open parenthesis minus 1, 8 close parenthesis$ , $open parenthesis 1, 2 close parenthesis$ , $open parenthesis 2, 5 close parenthesis$ , and $open parenthesis 3, 4 close parenthesis$ ?
$f (x) =$ _____

$f (x) = - x^{3} + 4 x^{2} - 2 x + 1$

Find the row reduced echelon form of the augmented matrix for the system
$\begin{aligned} - a & + & b & - & c & + & d & = & 8 \\ a & + & b & + & c & + & d & = & 2 \\ 8 a & + & 4 b & + & 2 c & + & d & = & 5 \\ 27 a & + & 9 b & - & 3 c & + & d & = & 4 \end{aligned}$
What third degree polynomial has a graph that passes through the points $open parenthesis minus 1, 8 close parenthesis$ , $open parenthesis 1, 2 close parenthesis$ , $open parenthesis 2, 5 close parenthesis$ , and $open parenthesis 3, 4 close parenthesis$ ?

$[\begin{array}{ccccc} 1 & 0 & 0 & 0 & - 1 \\ 0 & 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 & - 2 \\ 0 & 0 & 0 & 1 & 1 \end{array}]$
$f (x) = - x^{3} + 4 x^{2} - 2 x + 1$

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Matrix
Row echelon form
Upper triangular form
Coefficient matrix
Dimension
Entry
Reduced row echelon form
Elementary row operation
Augmented matrix
Square matrix
Order

CONCEPTS

We can use a matrix to represent a system of linear equations. Each row of the matrix consists of the coefficients in one of the equations of the system.
We operate on a matrix by using the elementary row operations.
Note
Elementary Row Operations
1. Multiply the entries of any row by a nonzero real number.
2. Add a constant multiple of one row to another row.
3. Interchange two rows.
We can solve a linear system by matrix reduction.
Note
Solving a Linear System by Matrix Reduction
1. Write the augmented matrix for the system.
2. Using elementary row operations, transform the matrix into an equivalent one in upper triangular form.
3. Use back-substitution to find the solution to the system.
Note
Strategy for Matrix Reduction
1. If the first entry in the first row is zero, interchange that row with another. Obtain zeros in the first entries of the second and third rows by adding suitable multiples of the first row to the second and third rows.
2. If the second entry of the second row is zero, interchange the second and third rows. Obtain a zero in the second entry of the third row by adding a suitable multiple of the second row to the third row.
To reduce larger matrices, we start with the first row and work our way along the diagonal, using row operations to obtain nonzero entries on the diagonal and zeros below the diagonal entry.

STUDY QUESTIONS

What are the coefficient matrix and the augmented matrix for a system of equations?
What does it mean for a matrix to be in upper triangular form?
State the three elementary row operations.
Describe a two-step procedure for reducing a $3 times 3$ matrix to upper triangular form.
Once the matrix is in upper triangular form, how do we complete the solution of the system?
Why is the reduced row echelon form of an augmented matrix convenient for solving a system?

SKILLS

Practice each skill in the Homework problems listed.

Perform elementary row operations: #1–8
Choose an elementry row operation to convert a matrix: #9–18
Solve a system by matrix reduction: #19–34
Solve a system using the reduced row echelon form of the augmented matrix: #35–40

Homework 8.3

In Problems 1–8, perform the given elementary row operation on the matrices.

Multiply $row 2$ by $minus 3$ :

$[\begin{array}{ccc} - 2 & 1 & 0 \\ 3 & - 1 & 2 \end{array}]$

$[\begin{array}{ccc} - 2 & 1 & 0 \\ - 9 & 3 & - 6 \end{array}]$

Multiply $row 1$ by $\frac{1}{4}$ :

$[ 2, 0, 3; minus 1, 5, 4 ]$

Add $2 open parenthesis row 1 close parenthesis$ to $row 2$ :

$[\begin{array}{ccc} 1 & - 3 & 6 \\ - 2 & 4 & - 1 \end{array}]$

$[ 1, 3, 6; 0, minus 2, 11 ]$

Add $- 3 (row 1)$ to $row 2$ :

$[\begin{array}{ccc} 1 & - 4 & 8 \\ 3 & - 2 & 10 \end{array}]$

Interchange $row 1$ and $row 3$ :

$[\begin{array}{cccc} 0 & - 3 & 2 & - 3 \\ 2 & 6 & - 1 & 3 \\ 1 & 0 & - 2 & 5 \end{array}]$

$[\begin{array}{cccc} 1 & 0 & - 2 & 5 \\ 2 & 6 & - 1 & 3 \\ 0 & - 3 & 2 & - 3 \end{array}]$

Interchange $row 2$ and $row 3$ :

$[\begin{array}{cccc} 1 & 6 & 0 & - 2 \\ 0 & 0 & 5 & - 10 \\ 0 & 3 & - 2 & 8 \end{array}]$

Add $- 4 (row 1)$ to $row 3$ :

$[\begin{array}{cccc} 1 & 2 & 1 & - 5 \\ 0 & 4 & - 2 & 3 \\ 4 & - 1 & 6 & - 8 \end{array}]$

$[\begin{array}{cccc} 1 & 2 & 1 & - 5 \\ 0 & 4 & - 2 & 3 \\ 0 & - 9 & 2 & 12 \end{array}]$

Add $2 open parenthesis row 2 close parenthesis$ to $row 3$ :

$[\begin{array}{cccc} 1 & - 7 & 5 & 2 \\ 0 & 1 & - 3 & - 1 \\ 0 & - 2 & - 3 & 4 \end{array}]$

In Problems 9–18, use row operations on the first matrix to form an equivalent matrix with the given entries.

$[\begin{array}{ccc} 1 & - 3 & 2 \\ 2 & 1 & 4 \end{array}] ⟶ [\begin{array}{ccc} 1 & - 3 & 2 \\ 0 & ? & ? \end{array}]$

$[ 1, minus 3, 2; 0, 7, 0 ]$

$[\begin{array}{ccc} - 2 & 3 & 0 \\ 4 & 1 & 6 \end{array}] ⟶ [\begin{array}{ccc} - 2 & 3 & 0 \\ 0 & ? & ? \end{array}]$

$[\begin{array}{ccc} 2 & 6 & - 4 \\ 5 & 3 & 1 \end{array}] ⟶ [\begin{array}{ccc} 2 & 6 & - 4 \\ ? & 0 & ? \end{array}]$

$[ 2, 6, minus 4; 4, 0, 3 ]$

$[\begin{array}{ccc} 6 & 4 & - 2 \\ - 1 & - 2 & - 3 \end{array}] ⟶ [\begin{array}{ccc} 6 & 4 & - 2 \\ ? & 0 & ? \end{array}]$

$[\begin{array}{cccc} 1 & - 2 & 2 & 1 \\ 2 & 3 & - 1 & 6 \\ 4 & 1 & - 3 & 3 \end{array}] ⟶ [\begin{array}{cccc} 1 & - 2 & 2 & 1 \\ 0 & ? & ? & ? \\ 0 & ? & ? & ? \end{array}]$

$[\begin{array}{cccc} 1 & - 2 & 2 & 1 \\ 0 & 7 & - 5 & 4 \\ 0 & 9 & - 11 & - 1 \end{array}]$

$[\begin{array}{cccc} 2 & - 1 & 3 & - 1 \\ - 4 & 0 & 4 & 5 \\ 6 & 2 & - 1 & - 2 \end{array}] ⟶ [\begin{array}{cccc} 2 & - 1 & 3 & - 1 \\ 0 & ? & ? & ? \\ 0 & ? & ? & ? \end{array}]$

$[\begin{array}{cccc} - 1 & 4 & 3 & 2 \\ 2 & - 2 & - 4 & 6 \\ 1 & 2 & 3 & - 3 \end{array}] ⟶ [\begin{array}{cccc} - 1 & 4 & 3 & 2 \\ ? & 0 & ? & ? \\ ? & 0 & ? & ? \end{array}]$

$[\begin{array}{cccc} - 1 & 4 & 3 & 2 \\ 3 & 0 & - 5 & 14 \\ - 3 & 0 & - 3 & 8 \end{array}]$

$[\begin{array}{cccc} 3 & - 2 & 4 & - 4 \\ 2 & 2 & 1 & 2 \\ - 1 & 1 & 5 & - 1 \end{array}] ⟶ [\begin{array}{cccc} 3 & - 2 & 4 & - 4 \\ ? & ? & 0 & ? \\ ? & ? & 0 & ? \end{array}]$

$[\begin{array}{cccc} - 2 & 1 & - 3 & - 2 \\ 4 & 2 & 0 & 2 \\ 6 & - 1 & 2 & 0 \end{array}] ⟶ [\begin{array}{cccc} - 2 & 1 & - 3 & - 2 \\ 0 & ? & ? & ? \\ 0 & 0 & ? & ? \end{array}]$

$[\begin{array}{cccc} - 2 & 1 & - 3 & - 2 \\ 0 & 4 & - 6 & - 2 \\ 0 & 0 & - 4 & - 5 \end{array}]$

$[\begin{array}{cccc} - 1 & 2 & 3 & 3 \\ 4 & 0 & 1 & - 6 \\ 2 & 2 & - 3 & - 2 \end{array}] ⟶ [\begin{array}{cccc} - 1 & 2 & 3 & 3 \\ 0 & ? & ? & ? \\ 0 & 0 & ? & ? \end{array}]$

In Problems 19–26, use matrix reduction on the augmented matrix to solve the system.

$\begin{aligned} x + 3 y & = 11 \\ 2 x - y & = 1 \end{aligned}$

$open parenthesis 2, 3 close parenthesis$

$\begin{aligned} x - 5 y & = 11 \\ 2 x + 3 y & = - 4 \end{aligned}$

$\begin{aligned} x - 4 y & = - 6 \\ 3 x + y & = - 5 \end{aligned}$

$open parenthesis minus 2, 1 close parenthesis$

$\begin{aligned} x + 6 y & = 14 \\ 5 x + 3 y & = - 4 \end{aligned}$

$\begin{aligned} 2 x + y & = 5 \\ 3 x - 5 y & = 14 \end{aligned}$

$open parenthesis 3, minus 1 close parenthesis$

$\begin{aligned} 3 x - 2 y & = 16 \\ 4 x + 2 y & = 12 \end{aligned}$

$\begin{aligned} x - y & = - 8 \\ x + 2 y & = 9 \end{aligned}$

$(\frac{- 7}{3}, \frac{17}{3})$

$\begin{aligned} 4 x - 3 y & = 16 \\ 2 x + y & = 8 \end{aligned}$

In Problems 27–34, use matrix reduction on the augmented matrix to solve the system.

$\begin{aligned} x & + & 3 y & - & z & = 5 \\ 3 x & - & y & + & 2 z & = 5 \\ x & + & y & + & 2 z & = 7 \end{aligned}$

$open parenthesis 1, 2, 2 close parenthesis$

$\begin{aligned} x & - & 2 y & + & 3 z & = - 11 \\ 2 x & + & 3 y & - & z & = 6 \\ 3 x & - & y & - & z & = 2 \end{aligned}$

$\begin{aligned} 2 x & - & y & + & z & = 5 \\ x & - & 2 y & - & 2 z & = 2 \\ 3 x & + & 3 y & - & z & = 4 \end{aligned}$

$(2, \frac{- 1}{2}, \frac{1}{2})$

$\begin{aligned} x & - & 2 y & - & 2 z & = 4 \\ 2 x & + & y & - & 3 z & = 7 \\ x & - & y & - & z & = 3 \end{aligned}$

$\begin{aligned} 2 x & - & y & - & z & = - 4 \\ x & + & y & + & z & = - 5 \\ x & + & 3 y & - & 4 z & = 12 \end{aligned}$

$(- 3, 1, - 3)$

$\begin{aligned} x & - & 2 y & - & 5 z & = 2 \\ 2 x & + & 3 y & + & z & = 11 \\ 3 x & - & y & - & z & = 11 \end{aligned}$

$\begin{aligned} 2 x - y & = 0 \\ 3 y + z & = 7 \\ 2 x + 3 z & = 1 \end{aligned}$

$(\frac{5}{4}, \frac{5}{2}, \frac{- 1}{2})$

$\begin{aligned} 3 x - z & = 7 \\ 2 x + y & = 6 \\ 3 y - z & = 7 \end{aligned}$

In Problems 35-40, solve the system by finding the reduced row echelon form of the augmented matrix.

$\begin{aligned} a & + & 2 b & - & c & + & 4 d & = 20 \\ - a & + & b & + & 4 c & + & 5 d & = 7 \\ 2 a & + & 2 b & + & 6 c & + & 7 d & = 22 \\ 3 a & + & 2 b & - & 3 c & + & 4 d & = 28 \end{aligned}$

$open parenthesis 3, 4, minus 1, 2 close parenthesis$

$\begin{aligned} 2 a & - & b & + & 3 c & - & d & = - 6 \\ 3 a & - & b & + & 2 c & - & 2 d & = - 37 \\ a & + & 4 c & + & d & = 43 \\ 4 a & + & 8 b & - & 9 c & + & 6 d & = 14 \end{aligned}$

$\begin{aligned} 6 a & + & 12 b & + & 18 c & + & 24 d & = 1 \\ 3 a & + & 4 c & = 3 \\ - 2 a & + & b & + & 3 d & = - 4 \\ 18 a & + & 6 b & + & 30 c & = 23 \end{aligned}$

$(\frac{5}{3}, \frac{4}{3}, \frac{- 1}{2}, \frac{- 2}{3})$

$\begin{aligned} a & - & b & - & c & - & d & = 1 \\ 12 a & - & 18 b & - & 6 d & = 13 \\ 3 a & - & 8 c & + & 6 d & = 3 \\ 24 a & - & 36 b & - & 18 c & + & 66 d & = 17 \end{aligned}$

$\begin{aligned} 2 a & + & 3 b & + & 4 c & + & d & + & 4 e & = 8 \\ 2 a & - & 5 b & - & 6 c & + & 4 d & + & 3 e & = 4 \\ 3 a & + & 3 b & + & 2 c & - & 3 d & + & 7 e & = - 13 \\ - 3 a & + & 4 b & - & c & + & d & - & e & = 6 \\ 4 a & - & 4 b & + & 3 c & + & 2 d & - & 2 e & = 1 \end{aligned}$

$(- 5, - 2, 3, 4, 2)$

$\begin{aligned} 3 a & + & 2 b & + & 5 c & + & 2 d & + & 4 e & = 6 \\ 4 a & + & b & + & 5 c & + & 2 d & + & 8 e & = 13 \\ 6 a & + & 3 b & + & 9 c & + & 3 d & + & 4 e & = 17 \\ 7 a & + & 8 b & + & 4 c & + & 7 e & = - 2 \\ 8 a & + & 9 c & + & d & + & 9 e & = 26 \end{aligned}$

For Problems 41–44, use a linear system of equations to find each polynomial with the given properties. Hint: The coefficients of the polynomial will be the unknowns of the system.

Find a third-degree polynomial $p (x)$ that satisfies $p (1) = 2, p (2) = 4, p (3) = 6,$ and $p open parenthesis 4 close parenthesis equals 1$ .
Find a third-degree polynomial $p (x)$ that satisfies $p (1) = 2, p (2) = 4, p (3) = 6,$ and $p open parenthesis 4 close parenthesis equals 5$ .

$p (x) = \frac{- 7}{6} x^{3} + 7 x^{2} - \frac{65}{6} x + 7$
$p (x) = \frac{- 1}{2} x^{3} + 3 x^{2} - \frac{7}{2} x + 3$

Find a third-degree polynomial $p (x)$ that satisfies $p (1) = 1, p (2) = 4, p (3) = 9,$ and $p open parenthesis 4 close parenthesis equals 5$ .
Find a third-degree polynomial $p (x)$ that satisfies $p (1) = 2, p (2) = 4, p (3) = 6,$ and $p open parenthesis 4 close parenthesis equals 10$ .

Find a fourth-degree polynomial $p (x)$ that satisfies $p (- 2) = - 45, p (- 1) = - 2, p (0) = 5, p (1) = 6$ and $p (2) = - 17$

$p (x) = - 2 x^{4} + x^{3} - x^{2} + 3 x + 5$

Find a fourth-degree polynomial $p (x)$ that satisfies $p (- 2) = 81, p (- 1) = 9, p (0) = 3, p (1) = 15$ and $p open parenthesis 2 close parenthesis equals 117$ .

Problems 45–48 consider inconsistent and dependent systems.

Suppose that the reduced row echelon form of an augmented matrix is

$[ 1, 0, 0, 0, 0; 0, 1, 0, 0, 0; 0, 0, 1, 0, 0; 0, 0, 0, 0, 1 ]$

To what equation does the last row of the matrix correspond?
Is there a solution to the system?

$0 equals 1$
No

Suppose that the reduced row echelon form of an augmented matrix is

$[ 1, 3, 0, 0, 0; 0, 0, 0, 0, 0; 0, 0, 0, 0, 1; 0, 0, 0, 0, 0 ]$

To what equation does the third row of the matrix correspond?
Is there a solution to the system?

Suppose that the reduced row echelon form of an augmented matrix is

$[ 1, 0, 0, 0, 9; 0, 1, 0, 0, 4; 0, 0, 1, 0, 2; 0, 0, 0, 0, 0 ]$

To what equation does the last row of the matrix correspond?
If the variables of the system are $a$ , $b$ , $c$ , and $d$ , what do you know about $a$ , $b$ , and $c$ ?
Is there more than one ordered four-tuple that makes all the equations true? (Hint: Does the value of $d$ affect any of the equations?)

$0 equals 0$
$a = 9, b = 4, c = 2$
Yes (infinitely many)

Suppose that the reduced row echelon form of an augmented matrix is

$[ 1, 2, 0, 0, 9; 0, 0, 1, 0, 4; 0, 0, 0, 1, 2; 0, 0, 0, 0, 0 ]$

To what equation does the last row of the matrix correspond?
If the variables of the system are $a$ , $b$ , $c$ , and $d$ , what do you know about $c$ and $d$ ?
Is there more than one ordered four-tuple that makes all the equations true? (Hint: If you specify any value of $b$ , will you be able to find values of $a$ , $c$ , and $d$ that make all the equations true?)

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.

8.3 Solving Linear Systems Using Matrices

Coefficient Matrix and Augmented Matrix of a System

Elementary Row Operations

Matrix Reduction

Reducing a 3 × 3 Matrix

Solving Larger Systems

Reduced Row Echelon Form

Section Summary

Vocabulary

CONCEPTS

STUDY QUESTIONS

SKILLS

Homework 8.3

Reducing a $3 times 3$ Matrix