Precalculus by Richard Wright

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Blessed are those who are persecuted because of righteousness, for theirs is the kingdom of heaven. Matthew‬ ‭5‬:‭10‬ ‭NIV‬‬‬‬‬‬‬

9-04 Inverse Matrices

Mr. Wright teaches the lesson.

Summary: In this section, you will:

Euros
Figure 1: Investment results. credit (Pixnio/Bicanski)

Fred is investing $5000 in two different accounts to stay diversified. One account pays 6% simple interest and the other pays 9% simple interest. How much should he invest in each account to earn an average of 8%? So far, this book has covered several ways to solve this problem such as writing two equations and using graphing, substitution, elimination, or Gauss-Jordan elimination. These methods work great, except there is a better way that actually works better on computers. This lesson will be about solving systems using an inverse matrix.

Before talking about inverse matrices, the identity matrix needs to be introduced. The identity matrix is the matrix equivalent of 1 and it given the name I. The identity matrix can be any size square as needed. The downward diagonal elements are all 1 and all other elements are 0. Here are some examples.

$$ \left[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right] \text{or} \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right] \text{or} \left[\begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{matrix}\right] $$

A square matrix multiplied by the identity matrix of the same size will result in the original matrix.

$$ A·I = A $$

An Inverse matrices multiplied with its square matrix will produce the identity matrix. Because the identity matrix is like 1, the inverse matrix is the multiplicative inverse of a square matrix. So, inverse matrices can be used to solve matrix equations by canceling out a matrix.

$$ A·A^{-1} = I $$

Inverse of a 2×2 Matrix

The inverse of a 2×2 matrix can be found with a formula.

If \(A = \left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), then

$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$

Example 1: Find the Inverse of a 2×2 Matrix

Find the inverse of \(A = \left[\begin{matrix} 3 & -1 \\ 0 & 2 \end{matrix}\right]\).

Solution

Comparing the matrix to \(\left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), a = 3, b = −1, c = 0, and d = 2. Plug those into the formula.

$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$

$$ A^{-1} = \frac{1}{3(2)-(-1)(0)} \left[\begin{matrix} 2 & -(-1) \\ -0 & 3 \end{matrix}\right] $$

$$ = \frac{1}{6} \left[\begin{matrix} 2 & 1 \\ 0 & 3 \end{matrix}\right] $$

$$ = \left[\begin{matrix} \frac{1}{3} & \frac{1}{6} \\ 0 & \frac{1}{2} \end{matrix}\right] $$

Try It 1

Find the inverse of \(B = \left[\begin{matrix} 2 & -2 \\ 1 & -3 \end{matrix}\right]\).

Answer

\(B^{-1} = \left[\begin{matrix} \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{2} \end{matrix}\right]\)

Find the Inverse of a Square Matrix

To find the inverse of a square matrix, A

  1. Create an augmented matrix with the identity matrix \([A : I]\)
  2. Use Gauss-Jordan Elimination to change the left side into the identity matrix. The right side will then be the inverse, \([I : A^{-1}]\)

Example 2: Find an Inverse Matrix

Find the inverse of \(A = \left[\begin{matrix} 2 & 0 & 1 \\ 0 & -1 & -3 \\ -2 & 4 & 3 \end{matrix}\right]\).

Solution

Create an augmented matrix with the identity matrix.

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ -2 & 4 & 3 & : & 0 & 0 & 1 \end{matrix}\right] $$

Use Gauss-Jordan Elimination to turn the left side into the identity matrix. There is already a 0 in the 1st column on the 2nd row, so go straight to getting rid of the −2 in the 3rd row by adding the 1st row to the 3rd row.

When showing your work on your assignment, you typically only show the steps with the red numbers in them.

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ \color{blue}{-2} & 4 & 3 & : & 0 & 0 & 1 \end{matrix}\right] $$

$$ \begin{matrix} \swarrow \\ \downarrow \\ +\rightarrow \end{matrix} \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ \color{red}{0} & \color{red}{4} & \color{red}{4} & : & \color{red}{1} & \color{red}{0} & \color{red}{1} \end{matrix}\right] $$

Now work down the 2nd column. Get rid of the 4 in the 3rd row by multiplying the 2nd row by 4 and add to the 3rd row.

$$ \begin{matrix} \quad \\ \swarrow ×4 \\ +\rightarrow \end{matrix} \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ 0 & \color{blue}{4} & 4 & : & 1 & 0 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & -3 & : & 0 & 1 & 0 \\ \color{red}{0} & \color{red}{0} & \color{red}{-8} & : & \color{red}{1} & \color{red}{4} & \color{red}{1} \end{matrix}\right] $$

The matrix is almost in row echelon form except for the leading nonzero entry in the 2nd and 3rd rows is not 1. If they were turned into a 1, then there would be fractions. However, fractions would make the remaining process a bit more of a nuisance, so they will be left for now. It is time to begin up, start getting zeros in the 3rd column and work from bottom up. Get rid of the −3 in the 2nd row by multiplying the 3rd row by −3 and adding to 8 times the 2nd row.

$$ \begin{matrix} \quad \\ +\rightarrow ×8 \\ \nwarrow ×\left(-3\right) \end{matrix} \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ 0 & -1 & \color{blue}{-3} & : & 0 & 1 & 0 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} 2 & 0 & 1 & : & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{-8} & \color{red}{0} & : & \color{red}{-3} & \color{red}{-4} & \color{red}{-3} \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

Continue working up the the 3rd column. Get rid of the 1 in the 1st row adding the 3rd row to 8 times the 1st row.

$$ \begin{matrix} +\rightarrow ×8 \\ \uparrow\qquad \\ \nwarrow \quad \end{matrix} \left[\begin{matrix} 2 & 0 & \color{blue}{1} & : & 1 & 0 & 0 \\ 0 & -8 & 0 & : & -3 & -4 & -3 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{16} & \color{red}{0} & \color{red}{0} & : & \color{red}{9} & \color{red}{4} & \color{red}{1} \\ 0 & -8 & 0 & : & -3 & -4 & -3 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

Now that the 3rd column is done, move up the 2nd column. However, there is already a 0 in the 1st row of the 2nd column, so now just multiply to get leading 1's.

$$ \begin{matrix} ×\frac{1}{16} \\ ×\left(-\frac{1}{8}\right) \\ ×\left(-\frac{1}{8}\right) \end{matrix} \left[\begin{matrix} 16 & 0 & 0 & : & 9 & 4 & 1 \\ 0 & -8 & 0 & : & -3 & -4 & -3 \\ 0 & 0 & -8 & : & 1 & 4 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{0} & : & \color{red}{\frac{9}{16}} & \color{red}{\frac{1}{4}} & \color{red}{\frac{1}{16}} \\ \color{purple}{0} & \color{purple}{1} & \color{purple}{0} & : & \color{purple}{\frac{3}{8}} & \color{purple}{\frac{1}{2}} & \color{purple}{\frac{3}{8}} \\ \color{green}{0} & \color{green}{0} & \color{green}{1} & : & \color{green}{-\frac{1}{8}} & \color{green}{-\frac{1}{2}} & \color{green}{-\frac{1}{8}} \end{matrix}\right] $$

The inverse is the right side.

$$ A^{-1} = \left[\begin{matrix} \frac{9}{16} & \frac{1}{4} & \frac{1}{16} \\ \frac{3}{8} & \frac{1}{2} & \frac{3}{8} \\ -\frac{1}{8} & -\frac{1}{2} & -\frac{1}{8} \end{matrix}\right] $$

Try It 2

Find the inverse of \(B = \left[\begin{matrix} 3 & 1 & 0 \\ 1 & 2 & -2 \\ 0 & 1 & 4 \end{matrix}\right]\).

Answer

\(B^{-1} = \left[\begin{matrix} \frac{5}{13} & -\frac{2}{13} & -\frac{1}{13} \\ -\frac{2}{13} & \frac{6}{13} & \frac{3}{13} \\ \frac{1}{26} & -\frac{3}{26} & \frac{5}{26} \end{matrix}\right]\).

Use an Inverse Matrix to Solve a System of Equations

Inverse matrices can be used to solve a system of equations. This is done to by writing the system of equations as a matrix equation. Matrix A is made of the coefficients, B is the constants, and X is the variables. This is complicated because dividing by a matrix is not defined, so instead multiply by the inverse.

$$ A·X = B $$

You cannot divide by a matrix, so multiply both sides by the inverse. Remember order is important when performing matrix multiplication. To get the inverse to multiply with matrix A, the inverse had to go to the far left, so the inverse also has to go to the left of the right side of the equation.

$$ \color{blue}{A^{-1}} ·A·X = \color{blue}{A^{-1}}·B $$

Simplify by remembering \(A^{-1}·A = I\).

$$ \color{blue}{I}·X = A^{-1}·B $$

Simplify by using \(I·X = X\).

$$ \color{blue}{X} = A^{-1}·B $$

In summary, if you start with \(A·X = B\), then the solution is \(X = A^{-1}·B\). For a system of equations, A was the coefficients and B was the constants. The solution is simply the inverse of the coefficient matrix multiplied with the constants matrix.

Use an Inverse Matrix to Solve a System of Equations
  1. Write the system as a matrix equation \(A·X = B\) where A is the coefficients, X is the variables, and B is the constants.
  2. Find the inverse of the coefficient matrix, \(A^{-1}\).
  3. The solution is \(X = A^{-1}·B\), just multiply the inverse with the constant matrix.

Example 3: Use an Inverse Matrix to Solve a System of Equations

Solve using an inverse matrix \(\left\{\begin{align} 2x - 6y &= 5 \\ 8x + 4y &= -1 \end{align}\right.\)

Solution

Write the system as a matrix equation. [coefficients] × [variables] = [constants]

$$ \left[\begin{matrix} 2 & -6 \\ 8 & 4 \end{matrix}\right] \left[\begin{matrix} x \\ y \end{matrix}\right] = \left[\begin{matrix} 5 \\ -1 \end{matrix}\right] $$

Find the inverse of the coefficient matrix, \(\left[\begin{matrix} 2 & -6 \\ 8 & 4 \end{matrix}\right]\). Comparing the matrix to \(\left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), a = 2, b = -6, c = 8, and d = 4. Plug those into the formula.

$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$

$$ A^{-1} = \frac{1}{2(4)-(-6)(8)} \left[\begin{matrix} 4 & 6 \\ -8 & 2 \end{matrix}\right] $$

$$ A^{-1} = \frac{1}{56} \left[\begin{matrix} 4 & 6 \\ -8 & 2 \end{matrix}\right] $$

Multiply the inverse matrix with the constant matrix, \(X = A^{-1}·B\). See lesson 9-03 for multiplying matrices.

$$ \left[\begin{matrix} x \\ y \end{matrix}\right] = \frac{1}{56} \left[\begin{matrix} 4 & 6 \\ -8 & 2 \end{matrix}\right] \left[\begin{matrix} 5 \\ -1 \end{matrix}\right] $$

$$ \left[\begin{matrix} x \\ y \end{matrix}\right] = \frac{1}{56} \left[\begin{matrix} 14 \\ -42 \end{matrix}\right] $$

$$ \left[\begin{matrix} x \\ y \end{matrix}\right] = \left[\begin{matrix} \frac{1}{4} \\ -\frac{3}{4} \end{matrix}\right] $$

The solution is \(\left(\frac{1}{4}, -\frac{3}{4}\right)\).

Try It 3

Solve using an inverse matrix \(\left\{\begin{align} 2x + 7y &= 67 \\ 3x - 4y &= -56 \end{align}\right.\)

Answer

(−5, 11)

Lesson Summary

Inverse of a 2×2 Matrix

The inverse of a 2×2 matrix can be found with a formula.

If \(A = \left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), then

$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$


Find the Inverse of a Square Matrix

To find the inverse of a square matrix, A

  1. Create an augmented matrix with the identity matrix \([A : I]\)
  2. Use Gauss-Jordan Elimination to change the left side into the identity matrix. The right side will then be the inverse, \([I : A^{-1}]\)

Use an Inverse Matrix to Solve a System of Equations
  1. Write the system as a matrix equation \(A·X = B\) where A is the coefficients, X is the variables, and B is the constants.
  2. Find the inverse of the coefficient matrix, \(A^{-1}\).
  3. The solution is \(X = A^{-1}·B\), just multiply the inverse with the constant matrix.

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Practice Exercises

  1. Multiply the matrices and determine if they are inverses. (Hint: The product should be the identity matrix.)
    \(\left[\begin{matrix} 2 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 2 & 0 \end{matrix}\right]\) and \(\left[\begin{matrix} 0.5 & 0.5 & 0 \\ 0 & 0 & 0.5 \\ 0 & -1 & 0 \end{matrix}\right]\)
  2. Find the inverse matrix.

  3. \(\left[\begin{matrix} 2 & -1 \\ 3 & 1 \end{matrix}\right]\)
  4. \(\left[\begin{matrix} 1 & 4 \\ -2 & 5 \end{matrix}\right]\)
  5. \(\left[\begin{matrix} 2 & -1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & -1 \end{matrix}\right]\)
  6. \(\left[\begin{matrix} 1 & 3 & -2 \\ 1 & -2 & 1 \\ 4 & 0 & 2 \end{matrix}\right]\)
  7. \(\left[\begin{matrix} 3 & 2 & -3 \\ 0 & -4 & 2 \\ -3 & 0 & 1 \end{matrix}\right]\)
  8. Use inverse matrices to solve the system of equations.

  9. \(\left\{\begin{align} 2x + 2y &= -2 \\ x - 5y &= 8 \end{align}\right.\)
  10. \(\left\{\begin{align} x - 3y &= -6 \\ 2x + y &= -5 \end{align}\right.\)
  11. \(\left\{\begin{align} 3x + 2y - 3z &= 4 \\ -4y + 2z &= -6 \\ -3x + z &= -2 \end{align}\right.\)
  12. Problem Solving

  13. Fred is investing $5000 in two different accounts to stay diversified. One account pays 6% simple interest and the other pays 9% simple interest. Use an inverse matrix to determine how much he should invest in each account to earn an average of 8%.
  14. Mixed Review

  15. (9-03) \(\left[\begin{matrix} 2 & 5 & -7 \end{matrix}\right] - 3 \left[\begin{matrix} -1 & 10 & 3 \end{matrix}\right]\)
  16. (9-03) \(\left[\begin{matrix} 2 & 1 & -2 \end{matrix}\right] \left[\begin{matrix} 1 & 0 \\ -1 & -2 \\ 0 & 5 \end{matrix}\right]\)
  17. (9-01) Add 2 times row 1 to row 3 and identify what it accomplished. \(\left[\begin{matrix} 1 & 3 & 0 & 2 \\ 0 & 3 & -2 & -5 \\ -2 & 8 & -6 & 2 \end{matrix}\right]\)
  18. (6-03) If \(\overset{\rightharpoonup}{m} = \langle 2, -1 \rangle\) and \(\overset{\rightharpoonup}{n} = \langle -3, 1 \rangle\), find \(2 \overset{\rightharpoonup}{m} - \overset{\rightharpoonup}{n}\).
  19. (5-04) Solve \(2 \sin x − \sqrt{2} = 0\) on the interval \(0 ≤ x < 2π\).

Answers

  1. Show work, yes they are inverses
  2. \(\left[\begin{matrix} \frac{1}{5} & \frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{matrix}\right]\)
  3. \(\left[\begin{matrix} \frac{5}{13} & -\frac{4}{13} \\ \frac{2}{13} & \frac{1}{13} \end{matrix}\right]\)
  4. \(\left[\begin{matrix} \frac{1}{2} & \frac{1}{6} & \frac{1}{6} \\ 0 & \frac{1}{3} & \frac{1}{3} \\ 0 & 0 & -1 \end{matrix}\right]\)
  5. \(\left[\begin{matrix} \frac{2}{7} & \frac{3}{7} & \frac{1}{14} \\ -\frac{1}{7} & -\frac{5}{7} & \frac{3}{14} \\ -\frac{4}{7} & -\frac{6}{7} & \frac{5}{14} \end{matrix}\right]\)
  6. \(\left[\begin{matrix} -\frac{1}{3} & -\frac{1}{6} & -\frac{2}{3} \\ -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -1 & -\frac{1}{2} & -1 \end{matrix}\right]\)
  7. \(\left(\frac{1}{2}, -\frac{3}{2}\right)\)
  8. (−3, 1)
  9. (1, 2, 1)
  10. $1666.67 at 6%, $3333.33 at 9%
  11. \(\left[\begin{matrix} 5 & -25 & -16 \end{matrix}\right]\)
  12. \(\left[\begin{matrix} 1 & -12 \end{matrix}\right]\)
  13. \(\left[\begin{matrix} 1 & 3 & 0 & 2 \\ 0 & 3 & -2 & -5 \\ 0 & 14 & -6 & 6 \end{matrix}\right]\); Made a 0 in the 3rd row.
  14. \(\langle 7, -3 \rangle\)
  15. \(\frac{π}{4}, \frac{3π}{4}\)