Algebra 2 by Richard Wright

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1-07 Use Inverse Matrices to Solve Linear Systems (12.4)

Mr. Wright teaches the lesson.

Objectives:

SDA NAD Content Standards (2018): AII.4.1, AII.6.1

Computer code
Figure 1: Computer code. (Pixabay/Elchinator)

Computer programming often use matrices called arrays. If a programmer needs to undo a matrix multiplication operation, then there is a problem because there is no matrix division! Instead the programmer would have to use an inverse matrix.

Identity Matrix

Matrices can be used to solve linear systems in a way that is different from Cramer's Rule. Before doing that; however, inverse matrices must be introduced.

Inverse matrices use the identity matrix. The identity matrix is a square matrix whose downward diagonals are 1's and the rest of the elements are 0's. It can be any sized square matrix. Examples are

$$ \left[\begin{matrix}1&0\\0&1\\\end{matrix}\right]\ \text{ and } \left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right] $$

The identity matrix multiplied with any matrix of the same dimension equals the original matrix. That means the identity matrix is the matrix equivalent of 1.

AI = IA = A

Division by a matrix is not defined. You cannot divide by a matrix! So, multiply by the inverse of a matrix instead. Any matrix times its inverse matrix equals the identity matrix.

A · A−1 = [1] = I

This is similar to how a function times its inverse also equals 1 like \(x\left(x^{-1}\right)\ = x\left(\frac{1}{x}\right) = 1\). This is called the identity property.

Inverse of a 2×2 Matrix

The rule for the inverse of a 2×2 matrix is

If \(A=\left[\begin{matrix}a&b\\c&d\\\end{matrix}\right]\), then \(A^{-1}=\frac{1}{\left|\begin{matrix}a&b\\c&d\\\end{matrix}\right|}\left[\begin{matrix}d&-b\\-c&a\\\end{matrix}\right]\)

Notice that the elements in the downward diagonal switched places and the upward diagonal changed signs.

Example 1

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

Solution

Fill in the formula.

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

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

Simplify the determinant part.

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

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

Distribute the \(\frac{1}{3}\) to the matrix.

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

Example 2

Find the inverse of \(\left[\begin{matrix}3&-2\\4&5\\\end{matrix}\right]\).

Solution

Fill in the formula.

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

$$ A^{-1}=\frac{1}{\left|\begin{matrix}3&-2\\4&5\\\end{matrix}\right|}\left[\begin{matrix}5&2\\-4&3\\\end{matrix}\right] $$

Simplify the determinant part.

$$ =\frac{1}{3\left(5\right)-4\left(-2\right)}\left[\begin{matrix}5&2\\-4&3\\\end{matrix}\right] $$

$$ =\frac{1}{23}\left[\begin{matrix}5&2\\-4&3\\\end{matrix}\right] $$

Distribute the \(\frac{1}{23}\) to the matrix.

$$ =\left[\begin{matrix}\frac{5}{23}&\frac{2}{23}\\-\frac{4}{23}&\frac{3}{23}\\\end{matrix}\right] $$

You can verify that two matrices are inverses by multiplying them together. The product should be the identity matrix.

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

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

$$ =\left[\begin{matrix}1&0\\0&1\\\end{matrix}\right] $$

Solve a Matrix Equation

Inverse matrices can be used to solve a matrix equation. Multiplying by the inverse matrix is used where division would normally be used.

If A, B, and X are matrices, and A · X = B, then

Multiply both sides by the inverse of matrix A.

A−1 · A · X = A−1 · B

The product of a matrix and its inverse is the identity matrix, I.

I · X = A-1 · B

The product of the identity matrix and any other matrix is equal to the other matrix. And the matrix equation is solved.

X = A-1 · B

Example 3

Solve the matrix equation \(\left[\begin{matrix}2&-1\\3&0\\\end{matrix}\right]X=\left[\begin{matrix}-1&0\\3&6\\\end{matrix}\right]\).

Solution

Start by finding the inverse of the first matrix \(\left[\begin{matrix}2&-1\\3&0\\\end{matrix}\right]\).

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

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

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

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

Multiply both sides by the inverse. Remember the order of matrix multiplication is important, so make sure the inverse goes in the same place on both sides of the equation.

A−1 A X = A−1 B

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

The A−1 A on the left cancels out. On the right, multiply the matrices.

$$ X=\left[\begin{matrix}0\left(-1\right)+\frac{1}{3}\left(3\right)&0\left(0\right)+\frac{1}{3}\left(6\right)\\-1\left(-1\right)+\frac{2}{3}\left(3\right)&-1\left(0\right)+\frac{2}{3}\left(6\right)\\\end{matrix}\right] $$

$$ X=\left[\begin{matrix}\mathbf{1}&\mathbf{2}\\\mathbf{3}&\mathbf{4}\\\end{matrix}\right] $$

Solve a System of Linear Equations

Solving a system of linear equations is similar to solving a matrix equation.

Use an Inverse Matrix to Solve a System of Linear Equations

To solve a system of linear equations using inverse matrices,

  1. Write the system as a matrix equation: [coefficient matrix] [variables] = [constants].
  2. Find the inverse of the coefficient matrix.
  3. Multiply both sides by the inverse. It cancels with the coefficient matrix.
  4. The result is [variables] = [inverse][constants]

Example 4

Use inverse matrices to solve \(\left\{ \begin{alignat}{3} x&+&3y&=&-5 \\ 2x&-&y&=&4 \end{alignat} \right.\).

Solution

Write the system as a matrix equation in the form [coefficient matrix] [variables] = [constants].

$$ \left[\begin{matrix}1&3\\2&-1\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}-5\\4\\\end{matrix}\right] $$

Find the inverse of the coefficient matrix.

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

$$ \left[\begin{matrix}1&3\\2&-1\\\end{matrix}\right]^{-1}=\frac{1}{\left|\begin{matrix}1&3\\2&-1\\\end{matrix}\right|}\left[\begin{matrix}-1&-3\\-2&1\\\end{matrix}\right] $$

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

$$ =\frac{1}{-7}\left[\begin{matrix}-1&-3\\-2&1\\\end{matrix}\right]=\left[\begin{matrix}\frac{1}{7}&\frac{3}{7}\\\frac{2}{7}&-\frac{1}{7}\\\end{matrix}\right] $$

Multiply the front of both sides by the inverse. Left side becomes \(\left[\begin{matrix}x\\y\\\end{matrix}\right]\), so just multiply the right side.

$$ \left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}\frac{1}{7}&\frac{3}{7}\\\frac{2}{7}&-\frac{1}{7}\\\end{matrix}\right]\left[\begin{matrix}-5\\4\\\end{matrix}\right] $$

$$ \left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}\frac{1}{7}\left(-5\right)+\frac{3}{7}\left(4\right)\\\frac{2}{7}\left(-5\right)+\left(-\frac{1}{7}\right)\left(4\right)\\\end{matrix}\right] $$

$$ \left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}1\\-2\\\end{matrix}\right] $$

The solution is (1, −2).

Practice Problem

Page 676 #1, 5, 9, 11, 13, 15, 17, 27, 29, 31, and Mixed Review = 15

    Mixed Review

  1. (1-06) Evaluate \(\left|\begin{matrix}1&5\\-2&-3\\\end{matrix}\right|\).
  2. (1-06) Find the area of the triangle with vertices (2, 1), (−2, 3), and (1, 4).
  3. (1-05) Simplify \(\left[\begin{matrix}2&1\\\end{matrix}\right]\left[\begin{matrix}-1&0\\-2&3\\\end{matrix}\right]\).
  4. (1-04) Simplify \(\left[\begin{matrix}2&1\\\end{matrix}\right]+[\begin{matrix}-4&3\\\end{matrix}]\).
  5. (1-03) Solve the system algebraically \(\left\{ \begin{alignat}{4} 2x&-&3y&+&z&=&5 \\ &&2y&-&z&=&-4 \\ x&+&2y&-&2z&=&-6 \end{alignat} \right.\).

Answers

  1. 7
  2. 5
  3. [−4 3]
  4. [−2 4]
  5. (0, −1, 2)