M - Inverse Matrices Lesson

Inverse Matrices

Before we discuss inverse matrices, let's talk about the determinant of a matrix. The determinant is the difference of the product of the two diagonals. We will calculate the determinant of a 2x2 matrix by hand, but for anything greater, we can use the calculator! We can only find the determinant of square matrices (remember a square matrix contains the same number of rows and columns)!

If equation image indicator $LaTeX: A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , then $LaTeX: \det\left(A\right)=\left|A\right|= \begin{vmatrix} a & b \\ c & d \end{vmatrix} =ad-cb$ $\det\left(A\right)=\left|A\right|= \begin{vmatrix} a & b \\ c & d \end{vmatrix} =ad-cb$

Watch this video to determine how to find the determinant of a 3 x 3 matrix.

Two matrices are considered inverses if, when multiplied (both ways), they create an identity matrix. An identity matrix is a square matrix with 1's on the main diagonal and 0's everywhere else.

equation image indicator $LaTeX: \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 &0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0& 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$ $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 &0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0& 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$

Determine whether equation image indicator $LaTeX: A= \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} \text{and}\: B= \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$ $A= \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} \text{and}\: B= \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$ are inverse matrices.

If AB = BA = I (the identity matrix), then A and B are inverse matrices.

Step 1: Find AB

equation image indicator $LaTeX: AB= \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} A= \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} = A= \begin{bmatrix} -3+4 & 6+(-6) \\ -2+2 & 4+(-3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ $AB= \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} A= \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} = A= \begin{bmatrix} -3+4 & 6+(-6) \\ -2+2 & 4+(-3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Step 2: Find BA

equation image indicator $LaTeX: AB= \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -3+4 & 2+(-2) \\ -6+6 & 4+(-3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ $AB= \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -3+4 & 2+(-2) \\ -6+6 & 4+(-3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Since AB = BA = I, we can confirm that equation image indicator $LaTeX: B=A^{-1}\:and\:A=B^{-1}$ $B=A^{-1}\:and\:A=B^{-1}$ .

Determine whether A and B are inverse matrices.

1. Problem: $LaTeX: A= \begin{bmatrix} -4 & 3 \\ 3 & -2 \end{bmatrix} \text{and}\:B= \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}$ $A= \begin{bmatrix} -4 & 3 \\ 3 & -2 \end{bmatrix} \text{and}\:B= \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}$ text annotation indicator

Solution: yes

2. Problem: $LaTeX: A= \begin{bmatrix} 6 & 2 \\ 2 & 1 \end{bmatrix} \text{and}\:B= \begin{bmatrix} 1 & -2 \\ -2 & 6 \end{bmatrix}$ $A= \begin{bmatrix} 6 & 2 \\ 2 & 1 \end{bmatrix} \text{and}\:B= \begin{bmatrix} 1 & -2 \\ -2 & 6 \end{bmatrix}$ text annotation indicator

Solution: no

So, let's try and find the inverse of a matrix by hand. Here is the formula:

equation image indicator $LaTeX: Let\:A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \text{then}\:A^{-1} =\frac{1}{ad-cb} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ $Let\:A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \text{then}\:A^{-1} =\frac{1}{ad-cb} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Notice that you are using the determinant in this formula!

If the determinant of a matrix is 0, then the inverse does not exist and the matrix is considered to be singular.

Watch this video to try an example:

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