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Solve Systems of Equations Using Matrices

Write the augmented matrix for a system of equations

Write the Augmented Matrix for a System of Equations

Solving a system of equations can be a tedious operation where a simple mistake can wreak havoc on finding the solution. An alternative method which uses the basic procedures of elimination but with notation that is simpler is available. The method involves using a matrix. A matrix is a rectangular array of numbers arranged in rows and columns.

We will use a matrix to represent a system of linear equations. We write each equation in standard form and the coefficients of the variables and the constant of each equation becomes a row in the matrix. Each column then would be the coefficients of one of the variables in the system or the constants. A vertical line replaces the equal signs. We call the resulting matrix the augmented matrix for the system of equations.

Notice the first column is made up of all the coefficients of x, the second column is the all the coefficients of y, and the third column is all the constants.

Example

Try it.

Write each system of linear equations as an augmented matrix:

ⓐ \(\{\begin{array}{l}5x-3y=-1 \\ y=2x-2\end{array}\) ⓑ \(\{\begin{array}{l}6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1\end{array}\)

Solution

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.

\[\begin{array}{lll}y & = & 2x-2 \\ -2x+y & = & -2\end{array}\]

We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Use Row Operations on a Matrix

Once a system of equations is in its augmented matrix form, we will perform operations on the rows that will lead us to the solution.

To solve by elimination, it doesn’t matter which order we place the equations in the system. Similarly, in the matrix we can interchange the rows.

When we solve by elimination, we often multiply one of the equations by a constant. Since each row represents an equation, and we can multiply each side of an equation by a constant, similarly we can multiply each entry in a row by any real number except 0.

In elimination, we often add a multiple of one row to another row. In the matrix we can replace a row with its sum with a multiple of another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

Performing these operations is easy to do but all the arithmetic can result in a mistake. If we use a system to record the row operation in each step, it is much easier to go back and check our work.

We use capital letters with subscripts to represent each row. We then show the operation to the left of the new matrix. To show interchanging a row:

Example

Try it.

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.

ⓑ Multiply row 2 by 5.

\[[\begin{array}{lllllllll}6 & & & -5 & & & 2 & & \\ 2 & & & 1 & & & -4 & & \\ 3 & & & -3 & & & 1 & & \end{array}|\begin{array}{lll} & & 3 \\ & & 5 \\ & & -1\end{array}]\]

Solution

ⓐ We interchange rows 2 and 3.

ⓑ We multiply row 2 by 5.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve Systems of Equations Using Matrices

To solve a system of equations using matrices, we transform the augmented matrix into a matrix in row-echelon form using row operations. 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.

Once we get the augmented matrix into row-echelon form, we can write the equivalent system of equations and read the value of at least one variable. We then substitute this value in another equation to continue to solve for the other variables. This process is illustrated in the next example.

How to Solve a System of Equations Using a Matrix

Try it.

Solve the system of equations using a matrix: \(\{\begin{array}{l}3x+4y=5 \\ x+2y=1\end{array}.\)

Solution

The steps are summarized here.

Here is a visual to show the order for getting the 1’s and 0’s in the proper position for row-echelon form.

We use the same procedure when the system of equations has three equations.

So far our work with matrices has only been with systems that are consistent and independent, which means they have exactly one solution. Let’s now look at what happens when we use a matrix for a dependent or inconsistent system.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Matrix: A matrix is a rectangular array of numbers arranged in rows and columns. A matrix with m rows and n columns has order \(m\ \times \ n.\) The matrix on the left below has 2 rows and 3 columns and so it has order \(2\ \times \ 3.\) We say it is a 2 by 3 matrix.

Each number in the matrix is called an element or entry in the matrix.
Row Operations: In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.
- Interchange any two rows
- Multiply a row by any real number except 0
- Add a nonzero multiple of one row to another row
Row-Echelon Form: 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.
How 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.

Solve Systems of Equations Using Matrices

Write the Augmented Matrix for a System of Equations

In the following exercises, write each system of linear equations as an augmented matrix.

Try it.

ⓐ \(\{\begin{array}{l}3x-y=-1 \\ 2y=2x+5\end{array}\)
ⓑ \(\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\)

Try it.

ⓐ \(\{\begin{array}{l}2x+4y=-5 \\ 3x-2y=2\end{array}\)
ⓑ \(\{\begin{array}{l}3x-2y-z=-2 \\ -2x+y=5 \\ 5x+4y+z=-1\end{array}\)

Solution

ⓐ \([\begin{array}{lllllll}2 & & & 4 & & & -5 \\ 3 & & & -2 & & & 2\end{array}]\)
ⓑ \([\begin{array}{llllllllll}3 & & & -2 & & & -1 & & & -2 \\ -2 & & & 1 & & & 0 & & & 5 \\ 5 & & & 4 & & & 1 & & & -1\end{array}]\)

Try it.

ⓐ \(\{\begin{array}{l}3x-y=-4 \\ 2x=y+2\end{array}\)
ⓑ \(\{\begin{array}{l}x-3y-4z=-2 \\ 4x+2y+2z=5 \\ 2x-5y+7z=-8\end{array}\)

Try it.

ⓐ \(\{\begin{array}{l}2x-5y=-3 \\ 4x=3y-1\end{array}\)
ⓑ \(\{\begin{array}{l}4x+3y-2z=-3 \\ -2x+y-3z=4 \\ \text{-}x-4y+5z=-2\end{array}\)

Solution

ⓐ \([\begin{array}{lllllll}2 & & & -5 & & & -3 \\ 4 & & & -3 & & & -1\end{array}]\)
ⓑ \([\begin{array}{llllllllll}4 & & & 3 & & & -2 & & & -3 \\ -2 & & & 1 & & & -3 & & & 4 \\ -1 & & & -4 & & & 5 & & & -2\end{array}]\)

Write the system of equations that corresponds to the augmented matrix.

Try it.

\([\begin{array}{llll}2 & & & -1 \\ 1 & & & -3\end{array}\ |\ \begin{array}{l}4 \\ 2\end{array}]\)

Try it.

\([\begin{array}{llll}2 & & & -4 \\ 3 & & & -3\end{array}\ |\ \begin{array}{l}-2 \\ -1\end{array}]\)

Solution

\(\{\begin{array}{l}2x-4y=-2 \\ 3x-3y=-1\end{array}\)

Try it.

\([\begin{array}{lllllll}1 & & & 0 & & & -3 \\ 1 & & & -2 & & & 0 \\ 0 & & & -1 & & & 2\end{array}\ |\ \begin{array}{l}-1 \\ -2 \\ 3\end{array}]\)

Try it.

\([\begin{array}{lllllll}2 & & & -2 & & & 0 \\ 0 & & & 2 & & & -1 \\ 3 & & & 0 & & & -1\end{array}\ |\ \begin{array}{l}-1 \\ 2 \\ -2\end{array}]\)

Solution

\(\{\begin{array}{l}2x-2y=-1 \\ 2y-z=2 \\ 3x-z=-2\end{array}\)

Use Row Operations on a Matrix

In the following exercises, perform the indicated operations on the augmented matrices.

Try it.

\([\begin{array}{llll}6 & & & -4 \\ 3 & & & -2\end{array}\ |\ \begin{array}{l}3 \\ 1\end{array}]\)

ⓐ Interchange rows 1 and 2

ⓑ Multiply row 2 by 3

Try it.

\([\begin{array}{llll}4 & & & -6 \\ 3 & & & 2\end{array}\ |\ \begin{array}{l}-3 \\ 1\end{array}]\)

ⓐ Interchange rows 1 and 2

ⓑ Multiply row 1 by 4

Solution

ⓐ \([\begin{array}{lllllll}3 & & & 2 & & & 1 \\ 4 & & & -6 & & & -3\end{array}]\)
ⓑ \([\begin{array}{lllllll}12 & & & 8 & & & 4 \\ 4 & & & -6 & & & -3\end{array}]\)
ⓒ \([\begin{array}{lllllll}12 & & & 8 & & & 4 \\ 24 & & & -10 & & & -5\end{array}]\)

Try it.

\([\begin{array}{lllllll}4 & & & -12 & & & -8 \\ 4 & & & -2 & & & -3 \\ -6 & & & 2 & & & -1\end{array}\ |\ \begin{array}{l}16 \\ -1 \\ -1\end{array}]\)

ⓐ Interchange rows 2 and 3

ⓑ Multiply row 1 by 4

Try it.

\([\begin{array}{lllllll}6 & & & -5 & & & 2 \\ 2 & & & 1 & & & -4 \\ 3 & & & -3 & & & 1\end{array}\ |\ \begin{array}{l}3 \\ 5 \\ -1\end{array}]\)

ⓐ Interchange rows 2 and 3

ⓑ Multiply row 2 by 5

Solution

ⓐ \(\left[\begin{array}{llll}6 & -5 & 2 & 3 \\ 3 & -3 & 1 & -1 \\ 2 & 1 & -4 & 5\end{array}\right]\)
ⓑ \(\left[\begin{array}{llll}6 & -5 & 2 & 3 \\ 15 & -15 & 5 & -5 \\ 2 & 1 & -4 & 5\end{array}\right]\)
ⓒ \(\left[\begin{array}{llll}2 & -7 & 10 & -7 \\ 15 & -15 & 5 & -5 \\ 2 & 1 & -4 & 5\end{array}\right]\)

Try it.

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \([\begin{array}{llll}1 & & & 2 \\ -3 & & & -4\end{array}\ |\ \begin{array}{l}5 \\ -1\end{array}].\)

Try it.

Perform the needed row operations that will get the first entry in both row 2 and row 3 to be zero in the augmented matrix: \([\begin{array}{lllllll}1 & & & -2 & & & 3 \\ 3 & & & -1 & & & -2 \\ 2 & & & -3 & & & -4\end{array}\ |\ \begin{array}{l}-4 \\ 5 \\ -1\end{array}].\)

Solution

\([\begin{array}{llllllllll}1 & & & -2 & & & 3 & & & -4 \\ 0 & & & 5 & & & -11 & & & 17 \\ 0 & & & 1 & & & -10 & & & 7\end{array}]\)

Solve Systems of Equations Using Matrices

In the following exercises, solve each system of equations using a matrix.

Try it.

\(\{\begin{array}{l}2x+y=2 \\ x-y=-2\end{array}\)

Try it.

\(\{\begin{array}{l}3x+y=2 \\ x-y=2\end{array}\)

Solution

\((1,-1)\)

Try it.

\(\{\begin{array}{l}\text{-}x+2y=-2 \\ x+y=-4\end{array}\)

Try it.

\(\{\begin{array}{l}-2x+3y=3 \\ x+3y=12\end{array}\)

Solution

\((3,3)\)

In the following exercises, solve each system of equations using a matrix.

Try it.

\(\{\begin{array}{l}2x-3y+z=19 \\ -3x+y-2z=-15 \\ x+y+z=0\end{array}\)

Try it.

\(\{\begin{array}{l}2x-y+3z=-3 \\ \text{-}x+2y-z=10 \\ x+y+z=5\end{array}\)

Solution

\((-2,5,2)\)

Try it.

\(\{\begin{array}{l}2x-6y+z=3 \\ 3x+2y-3z=2 \\ 2x+3y-2z=3\end{array}\)

Try it.

\(\{\begin{array}{l}4x-3y+z=7 \\ 2x-5y-4z=3 \\ 3x-2y-2z=-7\end{array}\)

Solution

\((-3,-5,4)\)

Try it.

\(\{\begin{array}{l}x+2z=0 \\ 4y+3z=-2 \\ 2x-5y=3\end{array}\)

Try it.

\(\{\begin{array}{l}2x+5y=4 \\ 3y-z=3 \\ 4x+3z=-3\end{array}\)

Solution

\((-3,2,3)\)

Try it.

\(\{\begin{array}{l}2y+3z=-1 \\ 5x+3y=-6 \\ 7x+z=1\end{array}\)

Try it.

\(\{\begin{array}{l}3x-z=-3 \\ 5y+2z=-6 \\ 4x+3y=-8\end{array}\)

Solution

\((-2,0,-3)\)

Try it.

\(\{\begin{array}{l}2x+3y+z=12 \\ x+y+z=9 \\ 3x+4y+2z=20\end{array}\)

Try it.

\(\{\begin{array}{l}x+2y+6z=5 \\ \text{-}x+y-2z=3 \\ x-4y-2z=1\end{array}\)

Solution

no solution

Try it.

\(\{\begin{array}{l}x+2y-3z=-1 \\ x-3y+z=1 \\ 2x-y-2z=2\end{array}\)

Try it.

\(\{\begin{array}{l}4x-3y+2z=0 \\ -2x+3y-7z=1 \\ 2x-2y+3z=6\end{array}\)

Solution

no solution

Try it.

\(\{\begin{array}{l}x-y+2z=-4 \\ 2x+y+3z=2 \\ -3x+3y-6z=12\end{array}\)

Try it.

\(\{\begin{array}{l}\text{-}x-3y+2z=14 \\ \text{-}x+2y-3z=-4 \\ 3x+y-2z=6\end{array}\)

Solution

\(\left(\frac{8}{5},-\frac{42}{5},-\frac{24}{5}\right)\)

Try it.

\(\{\begin{array}{l}x+y-3z=-1 \\ y-z=0 \\ \text{-}x+2y=1\end{array}\)

Try it.

\(\{\begin{array}{l}x+2y+z=4 \\ x+y-2z=3 \\ -2x-3y+z=-7\end{array}\)

Solution

infinitely many solutions \((x,y,z)\) where \(x=5z+2;y=-3z+1;z\) is any real number

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve: \(3(x+2)+4=4(2x-1)+9.\)
If you missed this problem, review .

Odkryj odpowiedź

\(x=1\)
Solve: \(0.25p+0.25(p+4)=5.20.\)
If you missed this problem, review .

Odkryj odpowiedź

\(p=8.4\)
Evaluate when \(x=-2\) and \(y=3\text{:}\ 2{x}^{2}-xy+3{y}^{2}.\)
If you missed this problem, review .

Odkryj odpowiedź

\(41\)
Write each system of linear equations as an augmented matrix:
ⓐ \(\{\begin{array}{l}5x-3y=-1 \\ y=2x-2\end{array}\) ⓑ \(\{\begin{array}{l}6x-5y+2z=3 \\ 2x+y-4z=5 \\ 3x-3y+z=-1\end{array}\)

Odkryj odpowiedź

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.
\[\begin{array}{lll}y & = & 2x-2 \\ -2x+y & = & -2\end{array}\]
We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.
Write each system of linear equations as an augmented matrix:
ⓐ \(\{\begin{array}{l}3x+8y=-3 \\ 2x=-5y-3\end{array}\) ⓑ \(\{\begin{array}{l}2x-5y+3z=8 \\ 3x-y+4z=7 \\ x+3y+2z=-3\end{array}\)

Odkryj odpowiedź

ⓐ \([\begin{array}{lllllll}3 & & & 8 & & & -3 \\ 2 & & & 5 & & & -3\end{array}]\)
ⓑ \([\begin{array}{llllllllll}2 & & & -5 & & & 3 & & & 8 \\ 3 & & & -1 & & & 4 & & & 7 \\ 1 & & & 3 & & & 2 & & & -3\end{array}]\)
Write each system of linear equations as an augmented matrix:
ⓐ \(\{\begin{array}{l}11x=-9y-5 \\ 7x+5y=-1\end{array}\) ⓑ \(\{\begin{array}{l}5x-3y+2z=-5 \\ 2x-y-z=4 \\ 3x-2y+2z=-7\end{array}\)

Odkryj odpowiedź

ⓐ \([\begin{array}{lllllll}11 & & & 9 & & & -5 \\ 7 & & & 5 & & & -1\end{array}]\)
ⓑ \([\begin{array}{llllllllll}5 & & & -3 & & & 2 & & & -5 \\ 2 & & & -1 & & & -1 & & & 4 \\ 3 & & & -2 & & & 2 & & & -7\end{array}]\)
Write the system of equations that corresponds to the augmented matrix:
\([\begin{array}{lllllllll}4 & & & -3 & & & 3 & & \\ 1 & & & 2 & & & -1 & & \\ -2 & & & -1 & & & 3 & & \end{array}|\begin{array}{lll} & & -1 \\ & & 2 \\ & & -4\end{array}].\)

Odkryj odpowiedź

We remember that each row corresponds to an equation and that each entry is a coefficient of a variable or the constant. The vertical line replaces the equal sign. Since this matrix is a \(4\ \times \ 3\), we know it will translate into a system of three equations with three variables.
Write the system of equations that corresponds to the augmented matrix: \([\begin{array}{llllllllll}1 & & & -1 & & & 2 & & & 3 \\ 2 & & & 1 & & & -2 & & & 1 \\ 4 & & & -1 & & & 2 & & & 0\end{array}].\)

Odkryj odpowiedź

\(\{\begin{array}{l}\ \ x-y+2z=3 \\ 2x+y-2z=1 \\ \ 4x-y+2z=0\end{array}\)
Write the system of equations that corresponds to the augmented matrix: \([\begin{array}{llllllllll}1 & & & 1 & & & 1 & & & 4 \\ 2 & & & 3 & & & -1 & & & 8 \\ 1 & & & 1 & & & -1 & & & 3\end{array}].\)

Odkryj odpowiedź

\(\{\begin{array}{l}\ \ x+y+z=4 \\ 2x+3y-z=8 \\ x+y-z=3\end{array}\)
Perform the indicated operations on the augmented matrix:
ⓐ Interchange rows 2 and 3.
ⓑ Multiply row 2 by 5.
ⓒ Multiply row 3 by \(-2\) and add to row 1.
\[[\begin{array}{lllllllll}6 & & & -5 & & & 2 & & \\ 2 & & & 1 & & & -4 & & \\ 3 & & & -3 & & & 1 & & \end{array}|\begin{array}{lll} & & 3 \\ & & 5 \\ & & -1\end{array}]\]

Odkryj odpowiedź

ⓐ We interchange rows 2 and 3.

ⓑ We multiply row 2 by 5.

ⓒ We multiply row 3 by \(-2\) and add to row 1.
Perform the indicated operations sequentially on the augmented matrix:
ⓐ Interchange rows 1 and 3.
ⓑ Multiply row 3 by 3.
ⓒ Multiply row 3 by 2 and add to row 2.
\([\begin{array}{lllllllll}5 & & & -2 & & & -2 & & \\ 4 & & & -1 & & & -4 & & \\ -2 & & & 3 & & & 0 & & \end{array}|\begin{array}{lll} & & -2 \\ & & 4 \\ & & -1\end{array}]\)

Odkryj odpowiedź

ⓐ \([\begin{array}{llllllllll}-2 & & & 3 & & & 0 & & & -1 \\ 4 & & & -1 & & & -4 & & & 4 \\ 5 & & & -2 & & & -2 & & & -2\end{array}]\)
ⓑ \([\begin{array}{llllllllll}-2 & & & 3 & & & 0 & & & 0 \\ 4 & & & -1 & & & -4 & & & 4 \\ 15 & & & -6 & & & -6 & & & -6\end{array}]\)
ⓒ \([\begin{array}{lllllllllllll}-2 & & & 3 & & & 0 & & & -1 & & & \\ 34 & & & -13 & & & -16 & & & -8 \\ 15 & & & -6 & & & -6 & & & -6 & & & \end{array}]\)
Perform the indicated operations on the augmented matrix:
ⓐ Interchange rows 1 and 2,
ⓑ Multiply row 1 by 2,
ⓒ Multiply row 2 by 3 and add to row 1.
\([\begin{array}{lllllllll}2 & & & -3 & & & -2 & & \\ 4 & & & 1 & & & -3 & & \\ 5 & & & 0 & & & 4 & & \end{array}|\begin{array}{lll} & & -4 \\ & & 2 \\ & & -1\end{array}]\)

Odkryj odpowiedź

ⓐ \([\begin{array}{llllllllll}4 & & & 1 & & & -3 & & & 2 \\ 2 & & & -3 & & & -2 & & & -4 \\ 5 & & & 0 & & & 4 & & & -1\end{array}]\)
ⓑ \([\begin{array}{llllllllll}8 & & & 2 & & & -6 & & & 4 \\ 2 & & & -3 & & & -2 & & & -4 \\ 5 & & & 0 & & & 4 & & & -1\end{array}]\)
ⓒ \([\begin{array}{llllllllll}14 & & & -7 & & & -12 & & & -8 \\ 2 & & & -3 & & & -2 & & & -4 \\ 5 & & & 0 & & & 4 & & & -1\end{array}]\)
Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \([\begin{array}{llllll}1 & & & -1 & & \\ 4 & & & -8 & & \end{array}|\begin{array}{lll} & & 2 \\ & & 0\end{array}].\)

Odkryj odpowiedź

To make the 4 a 0, we could multiply row 1 by \(-4\) and then add it to row 2.
Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \([\begin{array}{llllll}1 & & & -1 & & \\ 3 & & & -6 & & \end{array}|\begin{array}{lll} & & 2 \\ & & 2\end{array}].\)

Odkryj odpowiedź

\([\begin{array}{lllllll}1 & & & -1 & & & 2 \\ 0 & & & -3 & & & -4\end{array}]\)
Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \([\begin{array}{llllll}1 & & & -1 & & \\ -2 & & & -3 & & \end{array}|\begin{array}{lll} & & 3 \\ & & 2\end{array}].\)

Odkryj odpowiedź

\([\begin{array}{lllllll}1 & & & -1 & & & 3 \\ 0 & & & -5 & & & 8\end{array}]\)
Solve the system of equations using a matrix: \(\{\begin{array}{l}3x+4y=5 \\ x+2y=1\end{array}.\)

Odkryj odpowiedź
Solve the system of equations using a matrix: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)

Odkryj odpowiedź

The solution is \((4,-1).\)
Solve the system of equations using a matrix: \(\{\begin{array}{l}2x+y=-4 \\ x-y=-2\end{array}.\)

Odkryj odpowiedź

The solution is \((-2,0).\)

Solve the system of equations using a matrix: \(\{\begin{array}{l}3x+8y+2z=-5 \\ 2x+5y-3z=0 \\ x+2y-2z=-1\end{array}.\)

Odkryj odpowiedź


Write the augmented matrix for the equations.
Interchange row 1 and 3 to get the entry in row 1, column 1 to be 1.
Using row operations, get zeros in column 1 below the 1.

The entry in row 2, column 2 is now 1.
Continue the process until the matrix is in row-echelon form.

The matrix is now in row-echelon form.
Write the corresponding system of equations.
Use substitution to find the remaining variables.

Write the solution as an ordered pair or triple.
Check that the solution makes the original equations true.	We leave the check for you.

Solve the system of equations using a matrix: \(\{\begin{array}{l}2x-5y+3z=8 \\ 3x-y+4z=7 \\ x+3y+2z=-3\end{array}.\)

Odkryj odpowiedź

\((6,-1,-3)\)
Solve the system of equations using a matrix: \(\{\begin{array}{l}-3x+y+z=-4 \\ \text{-}x+2y-2z=1 \\ 2x-y-z=-1\end{array}.\)

Odkryj odpowiedź

\((5,7,4)\)

Solve the system of equations using a matrix: \(\{\begin{array}{l}x+y+3z=0 \\ x+3y+5z=0 \\ 2x+4z=1\end{array}.\)

Odkryj odpowiedź


Write the augmented matrix for the equations.
The entry in row 1, column 1 is 1.
Using row operations, get zeros in column 1 below the 1.

Continue the process until the matrix is in row-echelon form.
Multiply row 2 by 2 and add it to row 3.
At this point, we have all zeros on the left of row 3.
Write the corresponding system of equations.
Since \(0\ne 1\) we have a false statement. Just as when we solved a system using other methods, this tells us we have an inconsistent system. There is no solution.

Solve the system of equations using a matrix: \(\{\begin{array}{l}x-2y+2z=1 \\ -2x+y-z=2 \\ x-y+z=5\end{array}.\)

Odkryj odpowiedź

no solution
Solve the system of equations using a matrix: \(\{\begin{array}{l}3x+4y-3z=-2 \\ 2x+3y-z=-12 \\ x+y-2z=6\end{array}.\)

Odkryj odpowiedź

no solution

Solve the system of equations using a matrix: \(\{\begin{array}{l}x-2y+3z=1 \\ x+y-3z=7 \\ 3x-4y+5z=7\end{array}.\)

Odkryj odpowiedź


Write the augmented matrix for the equations.
The entry in row 1, column 1 is 1.
Using row operations, get zeros in column 1 below the 1.

Continue the process until the matrix is in row-echelon form.
Multiply row 2 by \(-2\) and add it to row 3.
At this point, we have all zeros in the bottom row.
Write the corresponding system of equations.
Since \(0=0\) we have a true statement. Just as when we solved by substitution, this tells us we have a dependent system. There are infinitely many solutions.
Solve for y in terms of z in the second equation.
Solve the first equation for x in terms of z.
Substitute \(y=2z+2.\)
Simplify.
Simplify.
Simplify.
The system has infinitely many solutions \((\frac{8}{5},-\frac{42}{5},-\frac{24}{5})\)

Solve the system of equations using a matrix: \(\{\begin{array}{l}x+y-z=0 \\ 2x+4y-2z=6 \\ 3x+6y-3z=9\end{array}.\)

Odkryj odpowiedź

infinitely many solutions \((x,y,z)\text{,}\) where \(x=z-3;y=3;z\) is any real number.
Solve the system of equations using a matrix: \(\{\begin{array}{l}x-y-z=1 \\ \text{-}x+2y-3z=-4 \\ 3x-2y-7z=0\end{array}.\)

Odkryj odpowiedź

infinitely many solutions \((x,y,z)\text{,}\) where \(x=5z-2;y=4z-3;z\) is any real number.
ⓐ \(\{\begin{array}{l}3x-y=-1 \\ 2y=2x+5\end{array}\)
ⓑ \(\{\begin{array}{l}4x+3y=-2 \\ x-2y-3z=7 \\ 2x-y+2z=-6\end{array}\)
ⓐ \(\{\begin{array}{l}2x+4y=-5 \\ 3x-2y=2\end{array}\)
ⓑ \(\{\begin{array}{l}3x-2y-z=-2 \\ -2x+y=5 \\ 5x+4y+z=-1\end{array}\)

Odkryj odpowiedź

ⓐ \([\begin{array}{lllllll}2 & & & 4 & & & -5 \\ 3 & & & -2 & & & 2\end{array}]\)
ⓑ \([\begin{array}{llllllllll}3 & & & -2 & & & -1 & & & -2 \\ -2 & & & 1 & & & 0 & & & 5 \\ 5 & & & 4 & & & 1 & & & -1\end{array}]\)
ⓐ \(\{\begin{array}{l}3x-y=-4 \\ 2x=y+2\end{array}\)
ⓑ \(\{\begin{array}{l}x-3y-4z=-2 \\ 4x+2y+2z=5 \\ 2x-5y+7z=-8\end{array}\)
ⓐ \(\{\begin{array}{l}2x-5y=-3 \\ 4x=3y-1\end{array}\)
ⓑ \(\{\begin{array}{l}4x+3y-2z=-3 \\ -2x+y-3z=4 \\ \text{-}x-4y+5z=-2\end{array}\)

Odkryj odpowiedź

ⓐ \([\begin{array}{lllllll}2 & & & -5 & & & -3 \\ 4 & & & -3 & & & -1\end{array}]\)
ⓑ \([\begin{array}{llllllllll}4 & & & 3 & & & -2 & & & -3 \\ -2 & & & 1 & & & -3 & & & 4 \\ -1 & & & -4 & & & 5 & & & -2\end{array}]\)
\([\begin{array}{llll}2 & & & -1 \\ 1 & & & -3\end{array}\ |\ \begin{array}{l}4 \\ 2\end{array}]\)
\([\begin{array}{llll}2 & & & -4 \\ 3 & & & -3\end{array}\ |\ \begin{array}{l}-2 \\ -1\end{array}]\)

Odkryj odpowiedź

\(\{\begin{array}{l}2x-4y=-2 \\ 3x-3y=-1\end{array}\)
\([\begin{array}{lllllll}1 & & & 0 & & & -3 \\ 1 & & & -2 & & & 0 \\ 0 & & & -1 & & & 2\end{array}\ |\ \begin{array}{l}-1 \\ -2 \\ 3\end{array}]\)
\([\begin{array}{lllllll}2 & & & -2 & & & 0 \\ 0 & & & 2 & & & -1 \\ 3 & & & 0 & & & -1\end{array}\ |\ \begin{array}{l}-1 \\ 2 \\ -2\end{array}]\)

Odkryj odpowiedź

\(\{\begin{array}{l}2x-2y=-1 \\ 2y-z=2 \\ 3x-z=-2\end{array}\)
\([\begin{array}{llll}6 & & & -4 \\ 3 & & & -2\end{array}\ |\ \begin{array}{l}3 \\ 1\end{array}]\)
ⓐ Interchange rows 1 and 2
ⓑ Multiply row 2 by 3
ⓒ Multiply row 2 by \(-2\) and add row 1 to it.
\([\begin{array}{llll}4 & & & -6 \\ 3 & & & 2\end{array}\ |\ \begin{array}{l}-3 \\ 1\end{array}]\)
ⓐ Interchange rows 1 and 2
ⓑ Multiply row 1 by 4
ⓒ Multiply row 2 by 3 and add row 1 to it.

Odkryj odpowiedź

ⓐ \([\begin{array}{lllllll}3 & & & 2 & & & 1 \\ 4 & & & -6 & & & -3\end{array}]\)
ⓑ \([\begin{array}{lllllll}12 & & & 8 & & & 4 \\ 4 & & & -6 & & & -3\end{array}]\)
ⓒ \([\begin{array}{lllllll}12 & & & 8 & & & 4 \\ 24 & & & -10 & & & -5\end{array}]\)
\([\begin{array}{lllllll}4 & & & -12 & & & -8 \\ 4 & & & -2 & & & -3 \\ -6 & & & 2 & & & -1\end{array}\ |\ \begin{array}{l}16 \\ -1 \\ -1\end{array}]\)
ⓐ Interchange rows 2 and 3
ⓑ Multiply row 1 by 4
ⓒ Multiply row 2 by \(-2\) and add to row 3.
\([\begin{array}{lllllll}6 & & & -5 & & & 2 \\ 2 & & & 1 & & & -4 \\ 3 & & & -3 & & & 1\end{array}\ |\ \begin{array}{l}3 \\ 5 \\ -1\end{array}]\)
ⓐ Interchange rows 2 and 3
ⓑ Multiply row 2 by 5
ⓒ Multiply row 3 by \(-2\) and add to row 1.

Odkryj odpowiedź

ⓐ \(\left[\begin{array}{llll}6 & -5 & 2 & 3 \\ 3 & -3 & 1 & -1 \\ 2 & 1 & -4 & 5\end{array}\right]\)
ⓑ \(\left[\begin{array}{llll}6 & -5 & 2 & 3 \\ 15 & -15 & 5 & -5 \\ 2 & 1 & -4 & 5\end{array}\right]\)
ⓒ \(\left[\begin{array}{llll}2 & -7 & 10 & -7 \\ 15 & -15 & 5 & -5 \\ 2 & 1 & -4 & 5\end{array}\right]\)
Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \([\begin{array}{llll}1 & & & 2 \\ -3 & & & -4\end{array}\ |\ \begin{array}{l}5 \\ -1\end{array}].\)

Symbols used here

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Systems of Equations Using Matrices

Write the augmented matrix for a system of equations
Use row operations on a matrix
Solve systems of equations using matrices
Interchange any two rows.
Multiply a row by any real number except 0.
Add a nonzero multiple of one row to another row.
Write the augmented matrix for the system of equations.
Using row operations get the entry in row 1, column 1 to be 1.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Spróbuj sam.

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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