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K12 LibreTexts

8.6 Augmented Matrices

  • Page ID
    1025
  • The reason why the rules for row reducing matrices are the same as the rules for eliminating coefficients when solving a system of equations is because you are essentially doing the same thing in each case.  When you write and rewrite the equation every time you end up writing down lots of extra information.  Matrices take care of this information by embedding it in the location of each entry.  How would you use matrices to write the following system of equations?

    \(5 x+y=6\)

    \(x+y=10\)

     

    HEADING Solving Systems of Equations with Augmented Matrices

    In order to represent a system as a matrix equation, first write all the equations in standard form so that the coefficients of the variables line up in columns.  Then copy down just the coefficients in a coefficient matrix array.  Next copy the variables in a variable matrix and the constants into a constant matrix.

    \(x+y+z=9\)

    \(x+2 y+3 z=22\)

    \(2 x+3 y+4 z=31\)

    \(\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4\end{array}\right] \cdot\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}9 \\ 22 \\ 31\end{array}\right]\)

    The reason why this works is because of the way matrix multiplication is defined.

    \(\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4\end{array}\right] \cdot\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 x+1 y+1 z \\ 1 x+2 y+3 z \\ 2 z+3 y+4 z\end{array}\right]=\left[\begin{array}{c}9 \\ 22 \\ 31\end{array}\right]\)

    Notice how putting brackets around the two matrices on the right does very little to hide the fact that this is just a regular system of 3 equations and 3 variables.

    Once you have your system represented as a matrix you can solve it using an augmented matrix.  An augmented matrix is two matrices that are joined together and operated on as if they were a single matrix.  In the case of solving a system, you need to augment the coefficient matrix and the constant matrix.  The vertical line indicates the separation between the coefficient matrix and the constant matrix.

    \(\left[\begin{array}{lll|c}1 & 1 & 1 & 9 \\ 1 & 2 & 3 & 22 \\ 2 & 3 & 4 & 31\end{array}\right]\)

    To solve, reduce the matrix to reduced row echelon form.

    \(\left[\begin{array}{lll|c}1 & 1 & 1 & 9 \\ 1 & 2 & 3 & 22 \\ 2 & 3 & 4 & 31\end{array}\right]\)

    \(R_{1} \cdot-1+R_{2} \rightarrow\left[\begin{array}{lll|c}1 & 1 & 1 & 9 \\ 0 & 1 & 2 & 13 \\ 2 & 3 & 4 & 31\end{array}\right]\)

    \(R_{1} \cdot-2+R_{3} \rightarrow\left[\begin{array}{lll|c}1 & 1 & 1 & 9 \\ 0 & 1 & 2 & 13 \\ 0 & 1 & 2 & 13\end{array}\right]\)

    \(R_{2} \cdot-1+R_{3} \rightarrow\left[\begin{array}{lll|c}1 & 1 & 1 & 9 \\ 0 & 1 & 2 & 13 \\ 0 & 0 & 0 & 0\end{array}\right]\)

    Because the last row is all 0's, this system is dependent. Therefore, there are an infinite number of solutions.

    Examples

    Example 1

    Earlier, you were asked how to write a system of equations as a matrix equation. If you were to write the system as a matrix equation, you could write:

    \(\begin{aligned} 5 x+y &=6 \\ x+y &=10 \\\left[\begin{array}{ll}5 & 1 \\ 1 & 1\end{array}\right] \cdot\left[\begin{array}{l}x \\ y\end{array}\right] &=\left[\begin{array}{c}6 \\ 10\end{array}\right] \end{aligned}\)

    Example 2

    Solve the following system using an augmented matrix.

    \(x+y+z=6\)

    \(x-y-z=-4\)

    \(x+2 y+3 z=14\)

     

    \(\left[\begin{array}{ccc|c}1 & 1 & 1 & 6 \\ 1 & -1 & -1 & -4 \\ 1 & 2 & 3 & 14\end{array}\right]\)

    \(R_{1} \cdot-1+R_{2} \rightarrow\left[\begin{array}{ccc|c}1 & 1 & 1 & 6 \\ 0 & -2 & -2 & -10 \\ 1 & 2 & 3 & 8\end{array}\right]\)

    \(R_{1} \cdot-1+R_{3} \rightarrow\left[\begin{array}{ccc|c}1 & 1 & 1 & 6 \\ 0 & -2 & -2 & -10 \\ 0 & 1 & 2 & 8\end{array}\right]\)

    \(R_{3} \cdot-1+R_{1} \rightarrow\left[\begin{array}{ccc|c}1 & 0 & -1 & -2 \\ 0 & -2 & -2 & -10 \\ 0 & 1 & 2 & 8\end{array}\right]\)

    \(R_{3} \cdot 3+R_{2} \rightarrow\left[\begin{array}{ccc|c}1 & 0 & -1 & -2 \\ 0 & 1 & 4 & 14 \\ 0 & 1 & 2 & 8\end{array}\right]\)

    \(R_{2} \cdot-1+R_{3} \rightarrow\left[\begin{array}{ccc|c}1 & 0 & -1 & -2 \\ 0 & 1 & 4 & 14 \\ 0 & 0 & -2 & -6\end{array}\right]\)

    \(R_{3} \div-2 \rightarrow\left[\begin{array}{ccc|c}1 & 0 & -1 & -2 \\ 0 & 1 & 4 & 14 \\ 0 & 0 & 1 & 3\end{array}\right]\)

    \(R_{3}+R_{1} \rightarrow\left[\begin{array}{ccc|c}1 & 0 & 0 & 1 \\ 0 & 1 & 4 & 14 \\ 0 & 0 & 1 & 3\end{array}\right]\)

    \(R_{3} \cdot-4+R_{2} \rightarrow\left[\begin{array}{lll|l}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3\end{array}\right]\)

    Every matrix can be interpreted as its own linear system.  The final augmented matrix can be interpreted as:

    \(1 x+0 y+0 z=1\)

    \(0 x+1 y+0 z=2\)

    \(0 x+0 y+1 z=3\)

    Which means \(x=1, y=2, z=3\).

    Example 3

    Solve the following system using augmented Matrices.

    \(w+x+z=11\)

    \(w+x=9\)

    \(x+y=7\)

    \(y+z=5\)

    While substitution would work in this problem, the idea is to demonstrate how augmented matrices will work even with larger matrices.

    \(\left[\begin{array}{llll|c}1 & 1 & 0 & 1 & 11 \\ 1 & 1 & 0 & 0 & 9 \\ 0 & 1 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 & 5\end{array}\right]\)

    Switch \(R_{2}, R_{3},\) and \(R_{4} \rightarrow\left[\begin{array}{cccc|c}1 & 1 & 0 & 1 & 11 \\ 0 & 1 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 & 5 \\ 1 & 1 & 0 & 0 & 9\end{array}\right]\)

    \(R_{1} \cdot-1+R_{4} \rightarrow\left[\begin{array}{cccc|c}1 & 1 & 0 & 1 & 11 \\ 0 & 1 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & -1 & -2\end{array}\right]\)

    \(R_{4} \cdot-1 \rightarrow\left[\begin{array}{llll|c}1 & 1 & 0 & 1 & 11 \\ 0 & 1 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 1 & 2\end{array}\right]\)

    \(R_{4} \cdot-1+R_{1} \rightarrow\left[\begin{array}{llll|l}1 & 1 & 0 & 0 & 9 \\ 0 & 1 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 1 & 2\end{array}\right]\)

    \(R_{4} \cdot-1+R_{3} \rightarrow\left[\begin{array}{llll|l}1 & 1 & 0 & 0 & 9 \\ 0 & 1 & 1 & 0 & 7 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 2\end{array}\right]\)

    \(R_{3} \cdot-1+R_{2} \rightarrow\left[\begin{array}{llll|l}1 & 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 2\end{array}\right]\)

    \(R_{2} \cdot-1+R_{1} \rightarrow\left[\begin{array}{llll|c}1 & 0 & 0 & 0 & 5 \\ 0 & 1 & 0 & 0 & 4 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 2\end{array}\right]\)

    Thus, \(w=5, x=4, y=3, z=2\)

    Example 4

    Use an augmented matrix to solve the following system.

    \(3 x+y=-15\)

    \(x+2 y=15\)

    The row reduction steps are not shown, only the initial and final augmented matrices.

    \(\left[\begin{array}{cc|c}3 & 1 & -15 \\ 1 & 2 & 15\end{array}\right] \rightarrow\left[\begin{array}{cc|c}1 & 0 & -9 \\ 0 & 1 & 12\end{array}\right]\)

    Example 5

    Use an augmented matrix to solve the following system.

    \(\begin{aligned}-a+b-c &=0 \\ 2 a-2 b-3 c &=25 \\ 3 a-4 b+3 c &=2 \end{aligned}\)

    The row reduction steps are not shown, only the initial and final augmented matrices.

    \(\left[\begin{array}{ccc|c}-1 & 1 & -1 & 0 \\ 2 & -2 & -3 & 25 \\ 3 & -4 & 3 & 2\end{array}\right] \rightarrow\left[\begin{array}{ccc|c}1 & 0 & 0 & 3 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -5\end{array}\right]\)

    Review

    Solve the following systems of equations using augmented matrices.  If one solution does not exist, explain why not.

    \[\begin{array}{c} 4 x-2 y &=-20 \\ x-3 y &=-15 \end{array}\]

    \[\begin{array}{c} 3 x+5 y &=33 \\ -x &-2 y=-13 \end{array}\]

    \[\begin{array}{c} x+4 y=11 \\ 3 x+12 y=33 \end{array}\]

    \[\begin{array}{c} -3 x+y=-7 \\ -x+4 y=5 \end{array}\]

    \[\begin{array}{c} 3 x+y &=6 \\ -6 x-2 y &=10 \end{array}\]

    \[\begin{array}{c} 2 x-y+z &=4 \\ 4 x+7 y-z &=38 \\ -x+3 y+2 z &=23 \end{array}\]

    \[\begin{array}{c} 4 x+y-z &=-16 \\ -3 x+4 y+z &=18 \\ x+y-3 z &=-17 \end{array}\]

    \[\begin{array}{c} 3 x+2 y-3 z &=7 \\ -x+5 y+2 z &=29 \\ x+2 y+z &=15 \end{array}\]

    \[\begin{array}{c} 2 x+y-2 z &=4 \\ -4 x-2 y+4 z &=-8 \\ 3 x+y-z &=5 \end{array}\]

    \[\begin{array}{c} -x+3 y+z=11 \\ 3 x+y+2 z=27 \\ 5 x-y-z=5 \end{array}\]

    \[\begin{array}{c} 3 x+2 y+4 z &=21 \\ -2 x+3 y+z &=-11 \\ x+2 y-3 z &=-3 \end{array}\]

    \[\begin{array}{c} -x+2 y-6 z=4 \\ 8 x+5 y+3 z=-8 \\ 2 x-4 y+12 z=5 \end{array}\]

    \[\begin{array}{c} 3 x+5 y+8 z &=37 \\ -6 x+3 y+z &=42 \\ x+3 y-2 z &=5 \end{array}\]

    \[\begin{array}{c} 4 x+y-6 z &=-38 \\ 2 x+7 y+8 z &=108 \\ -3 x+2 y-3 z &=-15 \end{array}\]

    \[\begin{array}{c} 6 x+3 y-2 z &=-22 \\ -4 x-2 y+4 z &=28 \\ 3 x+3 y+2 z &=7 \end{array}\]

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