Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
Blessed are those who are persecuted because of righteousness, for theirs is the kingdom of heaven. Matthew 5:10 NIV
Summary: In this section, you will:
SDA NAD Content Standards (2018): PC.6.4
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
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 \(A = \left[\begin{matrix} 3 & -1 \\ 0 & 2 \end{matrix}\right]\).
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.
$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$
$$ A^{-1} = \frac{1}{3(2)-(-1)(0)} \left[\begin{matrix} 2 & -(-1) \\ -0 & 3 \end{matrix}\right] $$
$$ = \frac{1}{6} \left[\begin{matrix} 2 & 1 \\ 0 & 3 \end{matrix}\right] $$
$$ = \left[\begin{matrix} \frac{1}{3} & \frac{1}{6} \\ 0 & \frac{1}{2} \end{matrix}\right] $$
Find the inverse of \(B = \left[\begin{matrix} 2 & -2 \\ 1 & -3 \end{matrix}\right]\).
Answer
\(B^{-1} = \left[\begin{matrix} \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{2} & -\frac{1}{2} \end{matrix}\right]\)
To find the inverse of a square matrix, A
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.
$$ A^{-1} = \left[\begin{matrix} \frac{9}{16} & \frac{1}{4} & \frac{1}{16} \\ \frac{3}{8} & \frac{1}{2} & \frac{3}{8} \\ -\frac{1}{8} & -\frac{1}{2} & -\frac{1}{8} \end{matrix}\right] $$
Find the inverse of \(B = \left[\begin{matrix} 3 & 1 & 0 \\ 1 & 2 & -2 \\ 0 & 1 & 4 \end{matrix}\right]\).
Answer
\(B^{-1} = \left[\begin{matrix} \frac{5}{13} & -\frac{2}{13} & -\frac{1}{13} \\ -\frac{2}{13} & \frac{6}{13} & \frac{3}{13} \\ \frac{1}{26} & -\frac{3}{26} & \frac{5}{26} \end{matrix}\right]\).
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.
A−1·A·X = A−1·B
Simplify by remembering A−1·A = I.
I·X = A−1·B
Simplify by using I·X = X.
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.
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 \(\left(\frac{1}{4}, -\frac{3}{4}\right)\).
Solve using an inverse matrix \(\left\{\begin{align} 2x + 7y &= 67 \\ 3x - 4y &= -56 \end{align}\right.\)
Answer
(−5, 11)
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
$$ A^{-1} = \frac{1}{ad-bc} \left[\begin{matrix} d & -b \\ -c & a \end{matrix}\right] $$
To find the inverse of a square matrix, A
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