4.2: Systems of Linear Equations: Matrices
Learning Objectives
- Write the Augmented Matrix of a System of Linear Equations.
- Write the System of Linear Equations from the Augmented Matrix.
- Perform Row Operations on Matrix.
- Solve a System of Linear Equations using Matrices.
Matrices
A matrix is a rectangular array of numbers that is usually named by a capital letter: [latex]A,B,C,[/latex] and so on. Each entry in a matrix is referred to as [latex]aij[/latex], such that [latex]i[/latex] represents the row and [latex]j[/latex] represents the column. Matrices are often referred to by their dimensions: [latex]m×n[/latex] indicating [latex]m[/latex] rows and [latex]n[/latex] columns.
[latex]\begin{array}{cccc}Column\;1&Column\;2 \cdots&Column\;j \cdots&Column\;n\end{array}[/latex]
[latex]\begin{array}{c}Row\;1\\Row\;2\\\vdots\\Row\;m\end{array}\begin{vmatrix}a_{11}& a_{12}& \cdots& a_{1j}& \cdots& a_{1n}\\a_{21}& a_{22}& \cdots& a_{2j}& \cdots& a_{2n}\\\vdots& \vdots& \cdots& \vdots& \cdots& \vdots\\a_{m1}& a_{m2}& \cdots& a_{mj}& \cdots& a_{mn}\end{vmatrix}[/latex]
In the matrix are referred to as a_mn meaning the [latex]m[/latex] rows and [latex]n[/latex] columns, its size or dimensions: [latex]𝑚×𝑛[/latex].
Augmented Matrix of a System of Linear Equations
An augmented matrix has a vertical bar that separates the columns into two groups. The coefficients of each variable are placed to the left of the vertical bar, and the constants are placed to the right. If a variable is missing, its coefficient is considered to be 0.
Example 4.2-1-1: Write the following systems as augmented matrices and identify the size of the matrix.
[latex]\left\{\begin{array}{l}5x-y+z=0\\4y+2z=1\\3x+y+z=-1\end{array}\right.[/latex]
Key
Example 4.2-1-1: Write the following systems as augmented matrices and identify the size of the matrix.
[latex]\left\{\begin{array}{l}5x-y+z=0\\4y+2z=1\\3x+y+z=-1\end{array}\right.[/latex]
[latex]\begin{bmatrix}5&-3&1&0\\0&4&2&1\\3&1&1&-1\end{bmatrix}[/latex]
This format is the format you will see in the calculator, such as TI-84. It has 3 rows and 4 columns; thus, the size is 3×4.
Mathematically, if we have a function and its equation, we use a vertical line to represent the equal sign.
[latex]\left[\begin{array}{ccc|c} 5 & -1 & 1 & 0 \\ 0 & 4 & 2 & 1 \\ 3 & 1 & 1 & -1 \end{array}\right][/latex]
[latex]\begin{array}{c}\uparrow\\\boxed{Vertical\;line\;represents\;equal\;sign}\end{array}[/latex]
Your Turn
Practice 4.2-1-1
System of Linear Equations from the Augmented Matrix
- Identify the variables.
- Assign a variable (usually x, y, z, etc.) to each column before the vertical bar.
- The column after the bar contains the constants (the right-hand side of the equations).
- Write each row as a linear equation.
- Multiply each entry in a row by its corresponding variable.
- Set the sum equal to the number on the right of the vertical bar.
[latex]\left[\begin{array}{cc|c} 2 & 3 & -1 \\ 4 & -2 & 0 \end{array}\right][/latex]
The number after the vertical line (on the right) is the value of the equation. Therefore, we only assume the variables x and y.
Step 1:
Column 1→ x
Column 2→ y
Step 2:
Write equation. [latex]\left\{\begin{array}{l}2+3y=-1\\4x-2y=0\end{array}\right.[/latex]
Example 4.2-2-1: Write system of linear equations from the augmented matrix.
[latex]\left[\begin{array}{ccc|c} 8 & -1 & 3 & 2 \\ 1 & 0 & 2 & 5 \\ 6 & -2 & 1 & -7 \end{array}\right][/latex]
Key
Example 4.2-2-1: Write system of linear equations from the augmented matrix.
[latex]\left[\begin{array}{ccc|c} 8 & -1 & 3 & 2 \\ 1 & 0 & 2 & 5 \\ 6 & -2 & 1 & -7 \end{array}\right][/latex]
Answer: [latex]\left\{\begin{array}{l}8x-y+3z=2\\x+2z=5\\6x-2y+z=-7\end{array}\right.[/latex]
Your Turn
Practice 4.2-2-1
Perform Row Operations on Matrix
Matrix Row Operations
The following row operations produce matrices that represent systems with the same solution set.
- Interchange any two rows.
- Replace a row by a nonzero multiple of that row.
- Replace a row by the sum of that row and a constant nonzero multiple of some other row.
Example 4.2-3-1: Use the matrix perform the indicated row operation.
[latex]\left[\begin{array}{ccc|c} 2 & 4 & 4 & -6 \\ 2 & 2 & -3 & 5 \\ 1 & 4 & 0 & 3 \end{array}\right][/latex]
- [latex]R_1\leftrightarrow R_3[/latex]
- [latex]\frac12R_2[/latex]
Key
Example 4.2-3-1: Use the matrix perform the indicated row operation.
[latex]\left[\begin{array}{ccc|c} 2 & 4 & 4 & -6 \\ 2 & 2 & -3 & 5 \\ 1 & 4 & 0 & 3 \end{array}\right][/latex]
Labeling the matrix row:
[latex]\begin{array}{c}R_1\\R_2\\R_3\end{array}\left[\begin{array}{ccc|c}2&4&4&-6\\2&2&-3&5\\1&4&0&3\end{array}\right][/latex]
- [latex]R_1\leftrightarrow R_3[/latex]
[latex]\left[\begin{array}{ccc|c}1&4&0&3\\2&2&-3&5\\2&4&4&-6\end{array}\right][/latex]
- [latex]\frac12R_2[/latex]
[latex]\left[\begin{array}{ccc|c}2&4&4&-6\\{\color[rgb]{0.0, 0.44, 0.73}\frac{\mathbf1}{\mathbf2}}\cdot2&{\color[rgb]{0.0, 0.44, 0.73}\frac{\mathbf1}{\mathbf2}}\cdot2&{\color[rgb]{0.0, 0.44, 0.73}\frac{\mathbf1}{\mathbf2}}\cdot-3&{\color[rgb]{0.0, 0.44, 0.73}\frac{\mathbf1}{\mathbf2}}\cdot5\\1&4&0&3\end{array}\right]\longrightarrow\left[\begin{array}{ccc|c}2&4&4&-6\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf1}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf2}&\color[rgb]{0.0, 0.44, 0.73}\frac{\boldsymbol-\mathbf3}{\mathbf2}&\color[rgb]{0.0, 0.44, 0.73}\frac{\mathbf5}{\mathbf2}\\1&4&0&3\end{array}\right][/latex]
Your Turn
Practice 4.2-3-1
Solve a System of Linear Equations using Matrices
The goal is to use row operations to turn the matrix into this special form:
[latex]\left[\begin{array}{ccc|c}1&0&0&a\\0&1&0&b\\0&0&1&c\end{array}\right][/latex]
This is called the reduced row-echelon form (RREF). It makes it easy to read the solution: means [latex]x=a[/latex], [latex]y=b[/latex] and [latex]z=c[/latex] so solution is [latex](a, b, c)[/latex]
No Solutions:
[latex]\left[\begin{array}{cc|c}1&a&0\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf1}\end{array}\right]\quad or\quad\left[\begin{array}{ccc|c}1&0&0&a\\0&1&0&b\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf1}\end{array}\right][/latex]
False Statement: No solution
Infinitely Many Solutions:
[latex]\left[\begin{array}{cc|c}1&a&0\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}\end{array}\right]\quad or\quad\left[\begin{array}{ccc|c}1&0&a&b\\0&1&c&d\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}\end{array}\right]\quad or\quad\left[\begin{array}{ccc|c}1&a&0&b\\0&0&1&c\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}\end{array}\right]\;\;\;or\quad\left[\begin{array}{ccc|c}1&a&b&c\\0&0&0&0\\{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}&{\color[rgb]{0.0, 0.44, 0.73}\mathbf0}\end{array}\right][/latex]
True Statement: No solution
Row Echelon VS Reduced Row Echelon
| Row Echelon | Reduced Row Echelon |
|---|---|
| [latex]\left[\begin{array}{ccc|c}1&a&b&d\\0&1&c&e\\0&0&1&f\end{array}\right][/latex] | [latex]\left[\begin{array}{ccc|c}1&0&0&a\\0&1&0&b\\0&0&1&c\end{array}\right][/latex] |
| I’m the mayor of Linear City—you can call me Dr. Linear. In my city, everything runs on linear knowledge. My role is to help citizens uncover linear concepts and earn the linear award badge. The more people master these ideas, the stronger our city grows. Let’s build something great together! | I’m the mayor of Linear City—you can call me Dr. Linear. In my city, everything runs on linear knowledge. My role is to help citizens uncover linear concepts and earn the linear award badge. The more people master these ideas, the stronger our city grows. Let’s build something great together! |
For example:
[latex]\left\{\begin{array}{l}5x-y+z=0\\4y+2z=1\\3x+y+z=-1\end{array}\right.[/latex]
| Row Echelon | Reduced Row Echelon |
|---|---|
| [latex]\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&2\\0&0&1&3\end{array}\right][/latex] | [latex]\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&2\\0&0&1&3\end{array}\right][/latex] |
| We need to use substitution to find to find [latex]x=1[/latex], [latex]y=2[/latex], [latex]z=3[/latex] |
We can read solution directly [latex]x=1[/latex], [latex]y=2[/latex], [latex]z=3[/latex] |
Example 4.2-4-1: Use the matrix solve x, y, z.
[latex]\left\{\begin{array}{l}x+y+z=6\\2x+y+3z=14\\x+2y+z=9\end{array}\right.[/latex]
Key
Example 4.2-4-1: Use the matrix solve x, y, z.
[latex]\left\{\begin{array}{l}x+y+z=6\\2x+y+3z=14\\x+2y+z=9\end{array}\right.[/latex]
Step 1: name rows
[latex]\begin{array}{c}R_1\\R_2\\R_3\end{array}\left[\begin{array}{ccc|c}{\color[rgb]{0.0, 0.44, 0.73}\mathbf1}&{\color[rgb]{0.0, 0.0, 1.0}\mathbf1}&{\color[rgb]{0.5, 0.0, 0.5}\mathbf1}&{\color[rgb]{0.76, 0.0, 0.0}\mathbf6}\\2&1&3&14\\1&4&1&9\end{array}\right][/latex]
Step 2: operation
New [latex]R_2=R_2-2R_1[/latex]
[latex]\left[\begin{array}{ccc|c}2-2\left({\color[rgb]{0.0, 0.44, 0.73}\mathbf1}\right)&1-2\left({\color[rgb]{0.0, 0.0, 1.0}\mathbf1}\right)&3-2\left({\color[rgb]{0.5, 0.0, 0.5}\mathbf1}\right)&{\color[rgb]{0.1, 0.1, 0.1}14}{\color[rgb]{0.1, 0.1, 0.1}-}{\color[rgb]{0.1, 0.1, 0.1}2}{\color[rgb]{0.1, 0.1, 0.1}\left({\color[rgb]{0.76, 0.0, 0.0}\mathbf6}\right)}\end{array}\right]\xrightarrow{\boxed{R_2-2R_1}}\overset{\boxed{New\;R_2}}{\left[\begin{array}{ccc|c}0&-1&1&2\end{array}\right]}[/latex]
New [latex]R_3=R_3-R_1[/latex]
[latex]\left[\begin{array}{ccc|c}1-\left({\color[rgb]{0.0, 0.44, 0.73}\mathbf1}\right)&2-\left({\color[rgb]{0.0, 0.0, 1.0}\mathbf1}\right)&1-\left({\color[rgb]{0.5, 0.0, 0.5}\mathbf1}\right)&{\color[rgb]{0.1, 0.1, 0.1}9}{\color[rgb]{0.1, 0.1, 0.1}-}{\color[rgb]{0.1, 0.1, 0.1}\left({\color[rgb]{0.76, 0.0, 0.0}\mathbf6}\right)}\end{array}\right]\xrightarrow{\boxed{R_3-R_1}}\overset{\boxed{New\;R_3}}{\left[\begin{array}{ccc|c}0&1&0&3\end{array}\right]}[/latex]
New Matrix:
[latex]\begin{array}{c}R_1\\New\;R_2\\New\;R_3\end{array}\left[\begin{array}{ccc|c}{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}6}\\0&-1&1&2\\0&1&0&3\end{array}\right][/latex]
[latex]New\;R_2\leftrightarrow\;New\;R_3[/latex]
[latex]\left[\begin{array}{ccc|c}{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}6}\\0&1&0&3\\0&-1&1&2\end{array}\right][/latex]
[latex]New\;R_3\ast=New\;R_3\;New\;R_2[/latex]
[latex]\left[\begin{array}{ccc|c}0+\left(0\right)&-1+\left(1\right)&1+\left(0\right)&{\color[rgb]{0.1, 0.1, 0.1}2}{\color[rgb]{0.1, 0.1, 0.1}+}{\color[rgb]{0.1, 0.1, 0.1}\left(3\right)}\end{array}\right]\xrightarrow{\boxed{New\;R_3+New\;R_2}}\overset{\boxed{New\;R_3^\ast}}{\left[\begin{array}{ccc|c}0&0&1&5\end{array}\right]}[/latex]
[latex]\left[\begin{array}{ccc|c}{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}1}&{\color[rgb]{0.1, 0.1, 0.1}6}\\0&1&0&3\\0&0&1&5\end{array}\right][/latex]
Substitute the values of y and z into the first equation to find x.
[latex]x+3+5=6[/latex]
[latex]x+8=6[/latex]
[latex]x=-2[/latex]
Thus final solution: [latex]x=-2[/latex], [latex]y=3[/latex], [latex]z= 5[/latex]
