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# Inverse Matrices

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Finding the Inverse of a Matrix

At your school book fair, you buy 3 paperback and 2 hardcovers. Your best friend buys 2 paperbacks and 4 hardcovers. What is the inverse of the matrix represented by this situation?

### Guidance

Recall that the multiplicative inverse of a real number is the reciprocal of the number and that the product of a number and its multiplicative inverse is the multiplicative identity , or 1. For example: $\frac{3}{7} \times \frac{7}{3}=1$ . Now we need to define a multiplicative identity and a multiplicative inverse for a square matrix. For real numbers, 1 is considered the identity because we can multiply any number, $a$ , by 1 and the result is $a$ . In other words, the value of the number does not change. For matrices, the multiplicative inverse of a square matrix will be a square matrix in which the values of the main diagonal are 1 and the remaining values are all zero. The following are examples of identity matrices.

$I= \begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix} \qquad I= \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} \qquad I= \begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}$

The products below illustrate how we can multiply a matrix by the identity and the result will be the original matrix.

Given:

$A= \begin{bmatrix} 2 & -1\\-3 & 5 \end{bmatrix}, \ AI=\begin{bmatrix} 2 & -1\\-3 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix}= \begin{bmatrix} 2(1)+(-1)(0) & 2(0)+(-1)(1)\\ (-3)(1)+5(0) & (-3)(0)+5(1) \end{bmatrix} = \begin{bmatrix} 2 & -1\\-3 & 5 \end{bmatrix}$

In fact, it does not matter which order we multiply by the identity matrix. In other words, $AI=IA=A$ .

Now that we have defined an identity matrix, we can determine an inverse matrix such that $A(A^{-1})=(A^{-1})A=I$ .

The formula for finding the Inverse of a $2 \times 2$ matrix is:

Given:

$A=\begin{bmatrix}a & b\\c & d \end{bmatrix}, \ A^{-1}= \frac{1}{det[A]}\begin{bmatrix}d & -b\\-c & a \end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}d & -b\\-c & a \end{bmatrix},$

where $ad-bc \ne 0$

$^*$ Note: If $ad-bc = 0$ or $det[A]=0$ , the matrix $A$ is called singular . The inverse of a singular matrix cannot be determined.

#### Example A

Find the inverse of the matrix $\begin{bmatrix} 1 & 2\\-3 & 7 \end{bmatrix}$ and verify that your result is the inverse.

Solution: First, use the formula above to find the inverse.

$\begin{bmatrix} 1 & 2\\-3 & 7 \end{bmatrix}^{-1} = \frac{1}{(1)(7)-(2)(-3)}\begin{bmatrix} 7 & -2\\3 & 1 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 7 & -2\\3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{7}{13} & -\frac{2}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}$

Now , To verify that this is indeed the inverse we must show that the product of the inverse and the original matrix is the identity matrix for a $2 \times 2$ matrix. It will be easier to find this product using the form of the inverse where the reciprocal of the determinant has not been distributed inside the matrix as shown below:

$\frac{1}{13} \begin{bmatrix}7 & 2\\-3 & 1\end{bmatrix} \begin{bmatrix}1 & -2\\3 & 7\end{bmatrix} = \frac{1}{13}\begin{bmatrix}(7)(1)+(-2)(-3) & (7)(-2)+(2)(7)\\(-3)(1)+(1)(3) & (-3)(-2)+(1)(7)\end{bmatrix} = \frac{1}{13}\begin{bmatrix}13 & 0\\0 & 13\end{bmatrix} = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$

#### Example B

Are $\begin{bmatrix} 4 & -3\\-3 & 2 \end{bmatrix}$ and $\begin{bmatrix} -2 & -3\\-3 & -4 \end{bmatrix}$ inverses?

Solution: If the matrices are inverses then their product will be the identity matrix.

$\begin{bmatrix} 4 & -3\\-3 & 2 \end{bmatrix} \begin{bmatrix} -2 & -3\\-3 & -4 \end{bmatrix}= \begin{bmatrix}(4)(-2)+(-3)(-3) & (4)(-3)+(-3)(-4)\\(-3)(-2)+(2)(-3) & (-3)(-3)+(2)(-4) \end{bmatrix} = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix}$

Since the product is the identity matrix, the matrices are inverses of one another.

#### Example C

Find the inverse of the matrix

$\begin{bmatrix} 4 & 6\\-2 & -3 \end{bmatrix}.$

Solution: Use the formula above to find the inverse.

$\begin{bmatrix} 4 & 6\\-2 & -3 \end{bmatrix}^{-1} = \frac{1}{(4)(-3)-(6)(-2)}\begin{bmatrix} -3 & -6\\2 & 4 \end{bmatrix} = \frac{1}{0}\begin{bmatrix} -3 & -6\\2 & 4 \end{bmatrix} \Rightarrow \ & \text{the inverse does not exist.}\\& \text{This matrix is singular.}$

Intro Problem Revisit The matrix that represents this situation is: $\begin{bmatrix} 3 & 2\\2 & 4 \end{bmatrix}$

Use the formula you learned in this lesson to find the inverse.

$\begin{bmatrix} 3 & 2\\2 & 4 \end{bmatrix}^{-1} = \frac{1}{[(3)(4)-(2)(2)]}\begin{bmatrix} 4 & -2\\-2 & 3 \end{bmatrix} = \frac{1}{8}\begin{bmatrix} 4 & -2\\-2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{4}\\ -\frac{1}{4} & \frac{3}{8} \end{bmatrix}$

### Vocabulary

Multiplicative Inverse for Real Numbers
The reciprocal of a number. The product of a number and its multiplicative inverse will always be the multiplicative identity.
Multiplicative Identity for Real Numbers
The number 1. Any number can be multiplied by 1 without changing the value.
Multiplicative Identity for Matrices
The square matrix that can be multiplied by any square matrix of the same size without changing the values of the original matrix. This matrix will contain all 1’s in the main diagonal and zeros elsewhere.
Multiplicative Inverse for Matrices
The square matrix, $A^{-1}$ , such that $A^{-1}A=AA^{-1}=I$ .

### Guided Practice

1. Are matrices $\begin{bmatrix} -\frac{1}{3} & 1\\-\frac{1}{2} & 2 \end{bmatrix}$ and $\begin{bmatrix} -12 & 6\\-3 & 2 \end{bmatrix}$ inverses of each other?

2. Find the inverse of

$\begin{bmatrix} 4 & 2\\10 & 5 \end{bmatrix}.$

3. Find the inverse of

$\begin{bmatrix} 3 & -4\\6 & -7 \end{bmatrix}.$

1. $\begin{bmatrix} -\frac{1}{3} & 1\\-\frac{1}{2} & 2 \end{bmatrix}\begin{bmatrix} -12 & 6\\-3 & 2 \end{bmatrix} = \begin{bmatrix} \left(-\frac{1}{3}\right)(-12)+(1)(-3) & \left(-\frac{1}{3}\right)(6)+(1)(2) \\\left(-\frac{1}{2}\right)(-12)+(2)(-3) & \left(-\frac{1}{2}\right)(6)+(2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix}.$

Yes, they are inverses.

2. $\begin{bmatrix} 4 & 2\\10 & 5 \end{bmatrix}^{-1} = \frac{1}{(4)(5)-(2)(10)}\begin{bmatrix} 5 & -2\\-10 & 4 \end{bmatrix} = \frac{1}{0} \begin{bmatrix} 5 & -2\\-10 & 4 \end{bmatrix} \Rightarrow & \ \text{inverse does not exist.}\\& \text{This matrix is singular.}$

3. $\begin{bmatrix} 3 & -4\\6 & -7 \end{bmatrix}^{-1} = \frac{1}{(3)(-7)-(-4)(6)}\begin{bmatrix} -7 & 4\\-6 & 3 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -7 & 4\\-6 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{7}{3} & \frac{4}{3}\\-2 & 1 \end{bmatrix}$

### Practice

Determine whether the following pairs of matrices are inverses of one another.

1. .
$\begin{bmatrix} 5 & -15\\3 & -10 \end{bmatrix}$ and $\begin{bmatrix} 2 & -3\\\frac{3}{5} & -1 \end{bmatrix}$
1. .
$\begin{bmatrix} 3 & 7\\1 & 2 \end{bmatrix}$ and $\begin{bmatrix} -2 & 7\\1 & -3 \end{bmatrix}$
1. .
$\begin{bmatrix} -5 & 8\\1 & 3 \end{bmatrix}$ and $\begin{bmatrix} 3 & -8\\-1 & -5 \end{bmatrix}$
1. .
$\begin{bmatrix} -5 & 4\\-9 & 7 \end{bmatrix}$ and $\begin{bmatrix} 7 & -4\\9 & -5 \end{bmatrix}$

Find the inverse of each matrix below, if it exists.

1. .
$\begin{bmatrix} -11 & 7\\-3 & 2 \end{bmatrix}$
1. .
$\begin{bmatrix} 6 & -3\\8 & -5 \end{bmatrix}$
1. .
$\begin{bmatrix} 1 & 2\\4 & 9 \end{bmatrix}$
1. .
$\begin{bmatrix} -2 & -3\\6 & 9 \end{bmatrix}$
1. .
$\begin{bmatrix} -2 & 4\\5 & 3 \end{bmatrix}$
1. .
$\begin{bmatrix} 0 & 1\\3 & 2 \end{bmatrix}$
1. .
$\begin{bmatrix} -4 & 7\\0 & 2 \end{bmatrix}$
1. .
$\begin{bmatrix} 2 & -6\\-6 & 18 \end{bmatrix}$
1. .
$\begin{bmatrix} 7 & 5\\14 & 10 \end{bmatrix}$
1. .
$\begin{bmatrix} -2 & 5\\-2 & 6 \end{bmatrix}$
1. For two 2x2 matrices, A and B, to be inverses of each other, what must be true of AB and BA ?