Solving a System of Linear Equations Using Matrices

We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.

Example 6: Solving a System of Linear Equations Using Matrices

Solve the system of linear equations using matrices.

\begin{matrix} \begin{array}{l} x - y + z = 8 \end{array} \\ 2 x + 3 y - z = - 2 \\ 3 x - 2 y - 9 z = 9 \end{matrix}

$\begin{array}{c}\begin{array}{l}\hfill \\ \hfill \\ x-y+z=8\hfill \end{array}\\ 2x+3y-z=-2\\ 3x - 2y - 9z=9\end{array}$

Solution

First, we write the augmented matrix.

[\begin{array}{rrr} 1 & - 1 & 1 \\ 2 & 3 & - 1 \\ 3 & - 2 & - 9 \end{array} | \begin{array}{r} 8 \\ - 2 \\ 9 \end{array}]

$\left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill 1\\ \hfill 2& \hfill 3& \hfill -1\\ \hfill 3& \hfill -2& \hfill -9\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 8\\ \hfill -2\\ \hfill 9\end{array}\right]$

Next, we perform row operations to obtain row-echelon form.

\begin{array}{rrrrr} - 2 R_{1} + R_{2} = R_{2} \to [\begin{array}{rrrrrr} 1 & - 1 & 1 \\ 0 & 5 & - 3 \\ 3 & - 2 & - 9 \end{array} | \begin{array}{rr} 8 \\ - 18 \\ 9 \end{array}] & - 3 R_{1} + R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & - 1 & 1 \\ 0 & 5 & - 3 \\ 0 & 1 & - 12 \end{array} | \begin{array}{rr} 8 \\ - 18 \\ - 15 \end{array}] \end{array}

$\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 3& \hfill & \hfill -2& \hfill & \hfill -9& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill 9\end{array}\right]& \hfill & \hfill & \hfill & \hfill -3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill -15\end{array}\right]\end{array}$

The easiest way to obtain a 1 in row 2 of column 1 is to interchange ${R}_{2}$ and ${R}_{3}$ .

Interchange R_{2} and R_{3} \to [\begin{array}{rrrrrrr} 1 & - 1 & 1 & 8 \\ 0 & 1 & - 12 & - 15 \\ 0 & 5 & - 3 & - 18 \end{array}]

$\text{Interchange}{R}_{2}\text{and}{R}_{3}\to \left[\begin{array}{rrrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill & \hfill 8\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill & \hfill -15\\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill & \hfill -18\end{array}\right]$

Then

\begin{array}{l} \begin{array}{rrrrr} - 5 R_{2} + R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & - 1 & 1 \\ 0 & 1 & - 12 \\ 0 & 0 & 57 \end{array} | \begin{array}{rr} 8 \\ - 15 \\ 57 \end{array}] & - \frac{1}{57} R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & - 1 & 1 \\ 0 & 1 & - 12 \\ 0 & 0 & 1 \end{array} | \begin{array}{rr} 8 \\ - 15 \\ 1 \end{array}] \end{array} \end{array}

$\begin{array}{l}\\ \begin{array}{rrrrr}\hfill -5{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 57& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 57\end{array}\right]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 1\end{array}\right]\end{array}\end{array}$

The last matrix represents the equivalent system.

\begin{array}{l} x - y + z = 8 \\ y - 12 z = - 15 \\ z = 1 \end{array}

$\begin{array}{l}\text{ }x-y+z=8\hfill \\ \text{ }y - 12z=-15\hfill \\ \text{ }z=1\hfill \end{array}$

Using back-substitution, we obtain the solution as $\left(4,-3,1\right)$ .

Example 7: Solving a Dependent System of Linear Equations Using Matrices

Solve the following system of linear equations using matrices.

\begin{array}{r} - x - 2 y + z = - 1 \\ 2 x + 3 y = 2 \\ y - 2 z = 0 \end{array}

$\begin{array}{r}\hfill -x - 2y+z=-1\\ \hfill 2x+3y=2\\ \hfill y - 2z=0\end{array}$

Solution

Write the augmented matrix.

[\begin{array}{rrr} - 1 & - 2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & - 2 \end{array} | \begin{array}{r} - 1 \\ 2 \\ 0 \end{array}]

$\left[\begin{array}{rrr}\hfill -1& \hfill -2& \hfill 1\\ \hfill 2& \hfill 3& \hfill 0\\ \hfill 0& \hfill 1& \hfill -2\end{array}\text{ }|\text{ }\begin{array}{r}\hfill -1\\ \hfill 2\\ \hfill 0\end{array}\right]$

First, multiply row 1 by $-1$ to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.

- R_{1} \to [\begin{array}{rrrrrrr} 1 & 2 & - 1 & 1 \\ 2 & 3 & 0 & 2 \\ 0 & 1 & - 2 & 0 \end{array}]

$-{R}_{1}\to \left[\begin{array}{rrrrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1& \hfill & \hfill 1\\ \hfill 2& \hfill & \hfill 3& \hfill & \hfill 0& \hfill & \hfill 2\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill & \hfill 0\end{array}\right]$

R_{2} \leftrightarrow R_{3} \to [\begin{array}{rrrrr} 1 & 2 & - 1 \\ 0 & 1 & - 2 \\ 2 & 3 & 0 \end{array} | \begin{array}{rr} 1 \\ 0 \\ 2 \end{array}]

${R}_{2}\leftrightarrow {R}_{3}\to \left[\begin{array}{rrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill 3& \hfill & \hfill 0\end{array}\text{ }|\begin{array}{rr}\hfill & \hfill 1\\ \hfill & \hfill 0\\ \hfill & \hfill 2\end{array}\right]$

- 2 R_{1} + R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & 2 & - 1 \\ 0 & 1 & - 2 \\ 0 & - 1 & 2 \end{array} | \begin{array}{rr} 1 \\ 0 \\ 0 \end{array}]

$-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill -1& \hfill & \hfill 2& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 1\\ \hfill & \hfill 0\\ \hfill & \hfill 0\end{array}\right]$

R_{2} + R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & 2 & - 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 0 \end{array} | \begin{array}{rr} 2 \\ 1 \\ 0 \end{array}]

${R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 2& \hfill & \hfill -1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 0& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 2\\ \hfill & \hfill 1\\ \hfill & \hfill 0\end{array}\right]$

The last matrix represents the following system.

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

$\begin{array}{l}\text{ }x+2y-z=1\hfill \\ \text{ }y - 2z=0\hfill \\ \text{ }0=0\hfill \end{array}$

We see by the identity $0=0$ that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for $y$ and substituting it into the first equation we can solve for $z$ in terms of $x$ .

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

$\begin{array}{l}\text{ }x+2y-z=1\hfill \\ \text{ }y=2z\hfill \\ \hfill \\ x+2\left(2z\right)-z=1\hfill \\ \text{ }x+3z=1\hfill \\ \text{ }z=\frac{1-x}{3}\hfill \end{array}$

Now we substitute the expression for $z$ into the second equation to solve for $y$ in terms of $x$ .

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

$\begin{array}{l}\text{ }y - 2z=0\hfill \\ \text{ }z=\frac{1-x}{3}\hfill \\ \hfill \\ y - 2\left(\frac{1-x}{3}\right)=0\hfill \\ \text{ }y=\frac{2 - 2x}{3}\hfill \end{array}$

The generic solution is $\left(x,\frac{2 - 2x}{3},\frac{1-x}{3}\right)$ .

Try It 5

Solve the system using matrices.

\begin{matrix} x + 4 y - z = 4 \\ 2 x + 5 y + 8 z = 15 \\ x + 3 y - 3 z = 1 \end{matrix}

$\begin{array}{c}x+4y-z=4\\ 2x+5y+8z=15\\ x+3y - 3z=1\end{array}$

Solution

Q & A

Can any system of linear equations be solved by Gaussian elimination?

Yes, a system of linear equations of any size can be solved by Gaussian elimination.

How To: Given a system of equations, solve with matrices using a calculator.

Save the augmented matrix as a matrix variable $\left[A\right],\left[B\right],\left[C\right]\text{,} \dots$ .
Use the ref( function in the calculator, calling up each matrix variable as needed.

Example 8: Solving Systems of Equations with Matrices Using a Calculator

Solve the system of equations.

\begin{array}{r} 5 x + 3 y + 9 z = - 1 \\ - 2 x + 3 y - z = - 2 \\ - x - 4 y + 5 z = 1 \end{array}

$\begin{array}{r}\hfill 5x+3y+9z=-1\\ \hfill -2x+3y-z=-2\\ \hfill -x - 4y+5z=1\end{array}$

Solution

Write the augmented matrix for the system of equations.

[\begin{array}{rrr} 5 & 3 & 9 \\ - 2 & 3 & - 1 \\ - 1 & - 4 & 5 \end{array} | \begin{array}{r} 5 \\ - 2 \\ - 1 \end{array}]

$\left[\begin{array}{rrr}\hfill 5& \hfill 3& \hfill 9\\ \hfill -2& \hfill 3& \hfill -1\\ \hfill -1& \hfill -4& \hfill 5\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 5\\ \hfill -2\\ \hfill -1\end{array}\right]$

On the matrix page of the calculator, enter the augmented matrix above as the matrix variable $\left[A\right]$ .

[A] = [\begin{array}{rrrrrrr} 5 & 3 & 9 & - 1 \\ - 2 & 3 & - 1 & - 2 \\ - 1 & - 4 & 5 & 1 \end{array}]

$\left[A\right]=\left[\begin{array}{rrrrrrr}\hfill 5& \hfill & \hfill 3& \hfill & \hfill 9& \hfill & \hfill -1\\ \hfill -2& \hfill & \hfill 3& \hfill & \hfill -1& \hfill & \hfill -2\\ \hfill -1& \hfill & \hfill -4& \hfill & \hfill 5& \hfill & \hfill 1\end{array}\right]$

Use the ref( function in the calculator, calling up the matrix variable $\left[A\right]$ .

ref ([A])

$\text{ref}\left(\left[A\right]\right)$

Evaluate.

\begin{array}{l} [\begin{array}{rrrr} 1 & \frac{3}{5} & \frac{9}{5} & \frac{1}{5} \\ 0 & 1 & \frac{13}{21} & - \frac{4}{7} \\ 0 & 0 & 1 & - \frac{24}{187} \end{array}] \to \begin{array}{l} x + \frac{3}{5} y + \frac{9}{5} z = - \frac{1}{5} \\ y + \frac{13}{21} z = - \frac{4}{7} \\ z = - \frac{24}{187} \end{array} \end{array}

$\begin{array}{l}\hfill \\ \left[\begin{array}{rrrr}\hfill 1& \hfill \frac{3}{5}& \hfill \frac{9}{5}& \hfill \frac{1}{5}\\ \hfill 0& \hfill 1& \hfill \frac{13}{21}& \hfill -\frac{4}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{24}{187}\end{array}\right]\to \begin{array}{l}x+\frac{3}{5}y+\frac{9}{5}z=-\frac{1}{5}\hfill \\ \text{ }y+\frac{13}{21}z=-\frac{4}{7}\hfill \\ \text{ }z=-\frac{24}{187}\hfill \end{array}\hfill \end{array}$

Using back-substitution, the solution is $\left(\frac{61}{187},-\frac{92}{187},-\frac{24}{187}\right)$ .

Example 9: Applying 2 × 2 Matrices to Finance

Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?

Solution

We have a system of two equations in two variables. Let $x=$ the amount invested at 10.5% interest, and $y=$ the amount invested at 12% interest.

\begin{array}{l} x + y = 12, 000 \\ 0.105 x + 0.12 y = 1, 335 \end{array}

$\begin{array}{l}\text{ }x+y=12,000\hfill \\ 0.105x+0.12y=1,335\hfill \end{array}$

As a matrix, we have

[\begin{array}{rr} 1 & 1 \\ 0.105 & 0.12 \end{array} | \begin{array}{r} 12, 000 \\ 1, 335 \end{array}]

$\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0.105& \hfill 0.12\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 12,000\\ \hfill 1,335\end{array}\right]$

Multiply row 1 by $-0.105$ and add the result to row 2.

[\begin{array}{rr} 1 & 1 \\ 0 & 0.015 \end{array} | \begin{array}{r} 12, 000 \\ 75 \end{array}]

$\left[\begin{array}{rr}\hfill 1& \hfill 1\\ \hfill 0& \hfill 0.015\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 12,000\\ \hfill 75\end{array}\right]$

Then,

\begin{array}{l} 0.015 y = 75 \\ y = 5, 000 \end{array}

$\begin{array}{l}0.015y=75\hfill \\ \text{ }y=5,000\hfill \end{array}$

So $12,000 - 5,000=7,000$ .

Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.

Example 10: Applying 3 × 3 Matrices to Finance

Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?

Solution

We have a system of three equations in three variables. Let $x$ be the amount invested at 5% interest, let $y$ be the amount invested at 8% interest, and let $z$ be the amount invested at 9% interest. Thus,

\begin{array}{l} x + y + z = 10, 000 \\ 0.05 x + 0.08 y + 0.09 z = 770 \\ 2 x - z = 0 \end{array}

$\begin{array}{l}\text{ }x+y+z=10,000\hfill \\ 0.05x+0.08y+0.09z=770\hfill \\ \text{ }2x-z=0\hfill \end{array}$

As a matrix, we have

[\begin{array}{rrr} 1 & 1 & 1 \\ 0.05 & 0.08 & 0.09 \\ 2 & 0 & - 1 \end{array} | \begin{array}{r} 10, 000 \\ 770 \\ 0 \end{array}]

$\left[\begin{array}{rrr}\hfill 1& \hfill 1& \hfill 1\\ \hfill 0.05& \hfill 0.08& \hfill 0.09\\ \hfill 2& \hfill 0& \hfill -1\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 10,000\\ \hfill 770\\ \hfill 0\end{array}\right]$

Now, we perform Gaussian elimination to achieve row-echelon form.

\begin{array}{l} \begin{array}{l} - 0.05 R_{1} + R_{2} = R_{2} \to [\begin{array}{rrrrrr} 1 & 1 & 1 \\ 0 & 0.03 & 0.04 \\ 2 & 0 & - 1 \end{array} | \begin{array}{rr} 10, 000 \\ 270 \\ 0 \end{array}] \end{array} \\ - 2 R_{1} + R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & 1 & 1 \\ 0 & 0.03 & 0.04 \\ 0 & - 2 & - 3 \end{array} | \begin{array}{rr} 10, 000 \\ 270 \\ - 20, 000 \end{array}] \\ \frac{1}{0.03} R_{2} = R_{2} \to [\begin{array}{rrrrrr} 0 & 1 & 1 \\ 0 & 1 & \frac{4}{3} \\ 0 & - 2 & - 3 \end{array} | \begin{array}{rr} 10, 000 \\ 9, 000 \\ - 20, 000 \end{array}] \\ 2 R_{2} + R_{3} = R_{3} \to [\begin{array}{rrrrrr} 1 & 1 & 1 \\ 0 & 1 & \frac{4}{3} \\ 0 & 0 & - \frac{1}{3} \end{array} | \begin{array}{rr} 10, 000 \\ 9, 000 \\ - 2, 000 \end{array}] \end{array}

$\begin{array}{l}\begin{array}{l}\hfill \\ -0.05{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 2& \hfill & \hfill 0& \hfill & \hfill -1& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill 0\end{array}\right]\hfill \end{array}\hfill \\ -2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 0.03& \hfill & \hfill 0.04& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 270\\ \hfill & \hfill -20,000\end{array}\right]\hfill \\ \frac{1}{0.03}{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 0& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill \frac{4}{3}& \hfill \\ \hfill 0& \hfill & \hfill -2& \hfill & \hfill -3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 9,000\\ \hfill & \hfill -20,000\end{array}\right]\hfill \\ 2{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill 1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill \frac{4}{3}& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill -\frac{1}{3}& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 10,000\\ \hfill & \hfill 9,000\\ \hfill & \hfill -2,000\end{array}\right]\hfill \end{array}$

The third row tells us $-\frac{1}{3}z=-2,000$ ; thus $z=6,000$ .

The second row tells us $y+\frac{4}{3}z=9,000$ . Substituting $z=6,000$ , we get

\begin{array}{r} y + \frac{4}{3} (6, 000) = 9, 000 \\ y + 8, 000 = 9, 000 \\ y = 1, 000 \end{array}

$\begin{array}{r}\hfill y+\frac{4}{3}\left(6,000\right)=9,000\\ \hfill y+8,000=9,000\\ \hfill y=1,000\end{array}$

The first row tells us $x+y+z=10,000$ . Substituting $y=1,000$ and $z=6,000$ , we get

\begin{array}{l} x + 1, 000 + 6, 000 = 10, 000 \\ x = 3, 000 \end{array}

$\begin{array}{l}x+1,000+6,000=10,000\hfill \\ \text{ }x=3,000\text{ }\hfill \end{array}$

The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.

Try It 6

A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.

Solution

Solving Systems with Gaussian Elimination