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# Solving Linear Systems Using Matrices and Technology

## Use a graphing calculator to evaluate matrix operations

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Solving Linear Systems Using Matrices and Technology

Customers of an ice cream shop are asking for more sizes, so the shop decides to test out medium and large cones. During its test, the shop sells 24 medium chocolate cones and 32 large chocolate cones. It also sells 40 medium vanilla cones and 60 X-large vanilla cones. The shop's chocolate sales for the day totaled $292 and its vanilla sales totaled$525. How much did the shop charge for a medium cone versus a large cone?

### Solving Linear Systems using Matrices

In this concept we will be using the calculator to find the inverse matrix and do matrix multiplication. We have already entered matrices into the calculator and multiplied matrices together on the calculator. To find the inverse of a matrix using the calculator we can call up the matrix by name and use the x1\begin{align*}x^{-1}\end{align*} key on the calculator to get the inverse of the matrix.

Now that we know how to get an inverse using the calculator, let’s look at solving the system. When we translate our system into a matrix equation we have the expression AX=B\begin{align*}AX=B\end{align*}. From the last concept, we know that X=A1B\begin{align*}X=A^{-1}B\end{align*}. We can enter the coefficient matrix into matrix A\begin{align*}A\end{align*} and the constants into matrix B\begin{align*}B\end{align*} and have the calculator find A1B\begin{align*}A^{-1}B\end{align*} for us.

Let's use the calculator to solve the following problems.

1. Find the inverse of the matrix:

[3152]\begin{align*}\begin{bmatrix} -3 & -5 \\ 1 & 2 \end{bmatrix}\end{align*}

First, enter the matrix into the calculator in matrix [A]\begin{align*}[A]\end{align*}. Now, 2nd\begin{align*}2^{nd}\end{align*} QUIT to return to the home screen. From here, go back into the MATRIX menu and select [A]\begin{align*}[A]\end{align*} under NAMES and press ENTER. You should see [A]\begin{align*}[A]\end{align*} in the home screen. Now press the x1\begin{align*}x^{-1}\end{align*} button to get [A]1\begin{align*}[A]^{-1}\end{align*} and press ENTER. The result is the inverse matrix

[2153].\begin{align*}\begin{bmatrix} -2 & -5 \\ 1 & 3 \end{bmatrix}.\end{align*}

\begin{align*}^*\end{align*}Note that the calculator can find the inverse of any square matrix.

1. Solve the system:

10x+6y6x+9y=11=15\begin{align*}10x+6y &= 11\\ -6x+9y &= -15\end{align*}

Let A=[10669]\begin{align*} A = \begin{bmatrix} 10 & 6\\ -6 & 9 \end{bmatrix}\end{align*} and B=[1115].\begin{align*} B = \begin{bmatrix} 11\\ -15 \end{bmatrix}.\end{align*} Enter these matrices into the calculator. Now, from the home screen, we can call up matrix A\begin{align*}A\end{align*} (from NAMES in the matrix menu) and use the x1\begin{align*}x^{-1}\end{align*} button to get [A]1\begin{align*}[A]^{-1}\end{align*} on the home screen. Finally we can call up matrix B\begin{align*}B\end{align*} from the matrix menu to get [A]1[B]\begin{align*}[A]^{-1}[B]\end{align*} and press ENTER. The result is

[3223].\begin{align*}\begin{bmatrix} \frac{3}{2}\\ \frac{-2}{3} \end{bmatrix}.\end{align*} The solution to the system is (32,23)\begin{align*}\left(\frac{3}{2},- \frac{2}{3} \right)\end{align*}.

1. Solve the system:

3x+11y9x33y=2=6\begin{align*}3x+11y &= 2\\ -9x-33y &= -6\end{align*}

Let A=[391133]\begin{align*} A = \begin{bmatrix} 3 & 11 \\ -9 & -33 \end{bmatrix}\end{align*} and B=[26].\begin{align*} B = \begin{bmatrix} 2 \\ -6 \end{bmatrix}.\end{align*} Enter these matrices into the calculator. Now, from the home screen, we can call up matrix A\begin{align*}A\end{align*}(from NAMES in the matrix menu) and use the x1\begin{align*}x^{-1}\end{align*} button to get [A]1\begin{align*}[A]^{-1}\end{align*} on the home screen. Finally we can call up matrix B\begin{align*}B\end{align*} from the matrix menu to get [A]1[B]\begin{align*}[A]^{-1}[B]\end{align*} and press ENTER. The result is ERR: SINGULAR MATRIX. Recall that a singular matrix is a matrix for which the inverse does not exist. This message tells us that there is no unique solution to this system. We must now use an alternative method to determine whether there are infinite solutions or no solution to the system. Using linear combination as shown below we can see that there are infinite solutions to this system.

3(3x+11y=2) 9x+33y=69x33y=69x33y=6  0=0\begin{align*}& 3(3x+11y=2) \quad \Rightarrow \quad \ 9x+33y=6\\ & -9x-33y=-6 \qquad \quad \underline{-9x-33y=-6 \; \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0=0\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the cost of a medium cone versus a large cone.

Solve the system using matrices on the calculator:

24x+32y40x+60y=292=525\begin{align*}24x+32y &= 292\\ 40x+60y &= 525\end{align*}

The shop charges $4.50 for a medium cone and$5.75 for a large cone.

#### Example 2

Find the inverse of 237412541\begin{align*}\begin{bmatrix} -2 & 4 & -5 \\ 3 & -1 & 4 \\ 7 & 2 & 1 \end{bmatrix}\end{align*} using your calculator.

2374125411=9532553135314533353325311537531053\begin{align*}\begin{bmatrix} -2 & 4 & -5 \\ 3 & -1 & 4 \\ 7 & 2 & 1 \end{bmatrix}^{-1}= \begin{bmatrix} -\frac{9}{53} & -\frac{14}{53} & \frac{11}{53} \\ \frac{25}{53} & \frac{33}{53} & -\frac{7}{53} \\ \frac{13}{53} & \frac{32}{53} & -\frac{10}{53} \end{bmatrix}\end{align*}

The calculator will convert all the values in the matrix to fractions. Press MATH and ENTER to choose 1:>FRAC\begin{align*}1: >FRAC\end{align*}.

Solve the following systems using matrices on the calculator.

#### Example 32x+3y−6x−7y=13=−21\begin{align*}2x+3y &= 13\\ -6x-7y &= -21\end{align*}

Let A=[2637]\begin{align*}A=\begin{bmatrix} 2 & 3 \\ -6 & -7 \end{bmatrix}\end{align*} and B=[1321].\begin{align*}B=\begin{bmatrix} 13 \\ -21 \end{bmatrix}.\end{align*} Using the calculator A1B=[79].\begin{align*}A^{-1}B=\begin{bmatrix} -7 \\ 9 \end{bmatrix}.\end{align*} The solution is (-7, 9).

#### Example 4

3x5y+z4x+3y2z7x+y3z=24=9=30\begin{align*}3x-5y+z &= -24\\ -4x+3y-2z &= 9\\ 7x+y-3z &= -30\end{align*}

Let A=347531123\begin{align*}A=\begin{bmatrix} 3 & -5 & 1 \\ -4 & 3 & -2 \\ 7 & 1 & -3 \end{bmatrix}\end{align*} and B=24930.\begin{align*}B=\begin{bmatrix} -24 \\ 9 \\ -30 \end{bmatrix}.\end{align*} Using the calculator A1B=257.\begin{align*}A^{-1}B=\begin{bmatrix} -2 \\ 5 \\ 7 \end{bmatrix}.\end{align*} The solution is (-2, 5, 7).

### Review

Use a calculator to find the inverse of each matrix below.

1. [2832]\begin{align*}\begin{bmatrix} 2 & -3 \\ 8 & 2 \end{bmatrix}\end{align*}
2. [6382]\begin{align*}\begin{bmatrix} -6 & 8 \\ -3 & 2 \end{bmatrix}\end{align*}
3. 534251106\begin{align*}\begin{bmatrix} 5 & 2 & -1 \\ 3 & 5 & 0 \\ -4 & 1 & 6 \end{bmatrix}\end{align*}

Solve the systems using matrices and a calculator.

1. 2x+y3x2y=1=3\begin{align*}2x+y &= -1\\ -3x-2y &= -3\end{align*}
2. \begin{align*}2x+10y &= 6\\ x+11y &= -3\end{align*}
3. \begin{align*}-3x+y &= 17\\ 2x-3y &= -23\end{align*}
4. \begin{align*}10x-5y &= 9\\ 2.5y &= 5x +4\end{align*}
5. \begin{align*}4x-3y &= -12\\ 7x &= 2y+5\end{align*}
6. \begin{align*}2x+5y &= 0\\ -4x+10y &= -12\end{align*}
7. \begin{align*}7x+2y+3z &= -12\\ -8x-y + 4z &= -6\\ -10x+3y+5z &= 1\end{align*}
8. \begin{align*}2x-4y+11z &= 18\\ -3x+5y-13z &= -25\\ 5x+10y+10z &= 5\end{align*}
9. \begin{align*}8x+y-4z &= 4\\ -2z-3y+4z &= -7\\ 4 x+5y-8z &= 11\end{align*}

For the following word problems, set up a system of linear equations to solve using matrices.

1. A mix of 1 lb almonds and 1.5 lbs cashews sells for $15.00. A mix of 2 lbs almonds and 1 lb cashews sells for$17.00. How much does each nut cost per pound?
2. Maria, Rebecca and Sally are selling baked goods for their math club. Maria sold 15 cookies, 20 brownies and 12 cupcakes and raised $23.50. Rebecca sold 22 cookies, 10 brownies and 11 cupcakes and raised$19.85. Sally sold 16 cookies, 5 brownies and 8 cupcakes and raised $13.30. How much did they charge for each type of baked good? 3. Justin, Mark and George go to the farmer’s market to buy fruit. Justin buys 2 lbs apples, 3 lbs pears and 1 lb of peaches for$13.24. Mark buys 1 lb apples, 2 lbs pears and 3 lbs peaches for $13.34. George buys 3 lbs apples, 1 lb pears and 2 lbs peaches for$14.04. How much does each fruit cost per pound?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 4.12.

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