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# Inverse Matrices

## Use a formula to identify the inverse of 2 X 2 matrices

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Inverse Matrices

Two numbers are multiplicative inverses if their product is 1.  Every number besides the number 0 has a multiplicative inverse.  For matrices, two matrices are inverses of each other if they multiply to be the identity matrix.

What kinds of matrices do not have inverses?

#### Guidance

Consider a matrix $A$  that has inverse $A^{-1}$ .  How do you find matrix $A^{-1}$  if you just have matrix $A$

$A=\begin{bmatrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1\end{bmatrix}, A^{-1}=?$

The answer is that you augment matrix $A$  with the identity matrix and row reduce.

$\begin{bmatrix}\begin{matrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1 \end{matrix} \left| \begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right.\end{bmatrix}$

The row reducing is demonstrated in Example A.  The right part of the augmented Matrix is the inverse matrix $A^{-1}$ .

$& \begin{bmatrix}\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{matrix} \left| \begin{matrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & 0\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{matrix}\right.\end{bmatrix}\\& A^{-1}=\begin{bmatrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & 0\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{bmatrix}$

Fractions are usually unavoidable when computing inverses.

One reason why inverses are so powerful is because they allow you to solve systems of equations with the same logic as you would solve a single linear equation.  Consider the following system based on the coefficients of matrix $A$  from above.

$x+2y+3z &= 96\\x+0y+z &= 36\\0x+2y-z &= -12$

By writing this system as a matrix equation you get:

$\begin{bmatrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}96\\36\\-12\end{bmatrix}\\A \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}96\\36\\-12\end{bmatrix}$

If this were a normal linear equation where you had a constant times the variable equals a constant, you would multiply both sides by the multiplicative inverse of the coefficient.  Do the same in this case.

$A^{-1} \cdot A \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} &= A^{-1} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}\\\begin{bmatrix}x\\y\\z\end{bmatrix} &= A^{-1} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}$

All that is left is for you to perform the matrix multiplication to get the solution.  See Example B.

Example A

Show the steps for finding the inverse matrix  $A$ from the guidance section.

Solution:

$\begin{bmatrix}\begin{matrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1 \end{matrix} \left| \begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & -I & \rightarrow\\& \rightarrow &\end{matrix} \begin{bmatrix}\begin{matrix}1 & 2 & 3\\0 & -2 & -2\\0 & 2 & -1 \end{matrix} \left| \begin{matrix}1 & 0 & 0\\-1 & 1 & 0\\0 & 0 & 1\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\& \rightarrow &\\\rightarrow & +II & \rightarrow\end{matrix}$

$\begin{bmatrix}\begin{matrix}1 & 2 & 3\\0 & -2 & -2\\0 & 0 & -3 \end{matrix} \left| \begin{matrix}1 & 0 & 0\\-1 & 1 & 0\\-1 & 1 & 1\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & \div (-2) & \rightarrow\\\rightarrow & \div (-3) & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}1 & 2 & 3\\0 & 1 & 1\\0 & 0 & 1 \end{matrix} \left| \begin{matrix}1 & 0 & 0\\\frac{1}{2} & -\frac{1}{2} & 0\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & -3III & \rightarrow\\\rightarrow & -III & \rightarrow\\& \rightarrow &\end{matrix}$

$\begin{bmatrix}\begin{matrix}1 & 2 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{matrix} \left| \begin{matrix}0 & 1 & 1\\\frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & -2II & \rightarrow\\& \rightarrow &\\& \rightarrow &\end{matrix} \begin{bmatrix}\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{matrix} \left| \begin{matrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{matrix}\right.\end{bmatrix}$

The matrix on the right is the inverse matrix $A^{-1}$ .

$A^{-1}=\begin{bmatrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{bmatrix}$

Example B

Solve the following system of equations using inverse matrices.

$\begin{bmatrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}96\\36\\-12\end{bmatrix}$

Solution:

$A \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}96\\36\\-12\end{bmatrix}\\A^{-1} \cdot A \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} &= A^{-1} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}\\\begin{bmatrix}x\\y\\z\end{bmatrix} &= A^{-1} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}\\\begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{bmatrix} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}\\\begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}-\frac{1}{3} \cdot 96+\frac{4}{3} \cdot 36+\frac{1}{3} \cdot (-12)\\\frac{1}{6} \cdot 96-\frac{1}{6} \cdot 36+\frac{1}{3} \cdot (-12)\\\frac{1}{3} \cdot 96-\frac{1}{3} \cdot 36-\frac{1}{3} \cdot (-12)\end{bmatrix}\\\begin{bmatrix}x\\y\\z\end{bmatrix} &= \begin{bmatrix}12\\6\\24\end{bmatrix}$

Example C

Find the inverse of the following matrix.

$\begin{bmatrix}1 & 6\\4 & 24\end{bmatrix}$

Solution:

$\begin{bmatrix}\begin{matrix}1 & 6\\4 & 24\end{matrix} \left| \begin{matrix}1 & 0\\0 & 1\\\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & -4I & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}1 & 6\\0 & 0\end{matrix} \left| \begin{matrix}1 & 0\\-4 & 1\\\end{matrix}\right.\end{bmatrix}$

This matrix is not invertible because its rows are not linearly independent.  To test to see if a square matrix is invertible, check whether or not the determinant is zero.  If the determinant is zero then the matrix is not invertible because the rows are not linearly independent.

Concept Problem Revisited

Non-square matrices do not generally have inverses.  Square matrices that have determinants equal to zero do not have inverses.

#### Vocabulary

Multiplicative inverses are two numbers or matrices whose product is one or the identity matrix.

#### Guided Practice

1. Confirm matrix  $A$ and $A^{-1}$  are inverses by computing  $A^{-1} \cdot A$ and $A \cdot A^{-1}$ .

$A=\begin{bmatrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1\end{bmatrix}, A^{-1}=\begin{bmatrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{bmatrix}$

2. Use a calculator to compute $A^{-1}$ , compute $A^{-1} \cdot A$ , compute $A \cdot A^{-1}$  and compute $A^{-1} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}$ .

3. The identity matrix happens to be its own inverse.  Find another matrix that is its own inverse.

1.

$A^{-1} \cdot A &= \begin{bmatrix}-\frac{1}{3} & \frac{4}{3} & \frac{1}{3}\\\frac{1}{6} & -\frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3}\end{bmatrix} \cdot \begin{bmatrix}1 & 2 & 3\\1 & 0 & 1\\0 & 2 & -1\end{bmatrix}=\\a_{11} &= -\frac{1}{3} \cdot 1+\frac{4}{3} \cdot 1+\frac{1}{3} \cdot 0=1\\a_{22} &= \frac{1}{6} \cdot 2-\frac{1}{6} \cdot 0+\frac{1}{3} \cdot 2=1\\a_{33} &= \frac{1}{3} \cdot 3-\frac{1}{3} \cdot 1-\frac{1}{3} (-1)=1$

Note that the rest of the entries turn out to be zero.  This is left for you to confirm.

2. Start by entering just matrix $A$  into the calculator.

To compute matrix $A^{-1}$  use the inverse button programmed into the calculator.  Do not try to raise the matrix to the negative one exponent.  This will not work.

Note that the calculator may return decimal versions of the fractions and will not show the entire matrix on its limited display.  You will have to scroll to the right to confirm that $A^{-1}$  matches what you have already found.  Once you have found $A^{-1}$  go ahead and store it as matrix $B$  so you do not need to type in the entries.

$A^{-1} \cdot A=B \cdot A$

$A \cdot A^{-1}=A \cdot B$

$A^{-1} \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}=B \cdot \begin{bmatrix}96\\36\\-12\end{bmatrix}=B \cdot C$

You need to create matrix $C=\begin{bmatrix}96\\36\\-12\end{bmatrix}$

Being able to effectively use a calculator should improve your understanding of matrices and allow you to check all the work you do by hand.

3. Helmert came up with a very clever matrix that happens to be its own inverse.  Here are the $2 \times 2$  and the $3 \times 3$  versions.

$\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix}, \begin{bmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} &-\frac{2}{\sqrt{6}}\end{bmatrix}$

#### Practice

Find the inverse of each of the following matrices, if possible.  Make sure to do some by hand and some with your calculator.

1. $\begin{bmatrix}4 & 5\\2 & 3\end{bmatrix}$

2.  $\begin{bmatrix}-3 & 6\\2 & 5\end{bmatrix}$

3.  $\begin{bmatrix}-1 & 2\\2 & 0\end{bmatrix}$

4.  $\begin{bmatrix}1 & 6\\0 & 1\end{bmatrix}$

5.  $\begin{bmatrix}6 & 5\\2 & -2\end{bmatrix}$

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

7.  $\begin{bmatrix}-1 & 3 & -4\\4 & 2 & 1\\1 & 2 & 5\end{bmatrix}$

8.  $\begin{bmatrix}4 & 5 & 8\\9 & 0 & 1\\0 & 3 & -2\end{bmatrix}$

9.  $\begin{bmatrix}0 & 7 & -1\\2 & -3 & 1\\6 & 8 & 0\end{bmatrix}$

10.  $\begin{bmatrix}4 & 2 & -3\\2 & 4 & 5\\1 & 8 & 0\end{bmatrix}$

11.  $\begin{bmatrix}-2 & -6 & -12\\-1 & -5 & -2\\2 & 3 & 4\end{bmatrix}$

12.  $\begin{bmatrix}-2 & 6 & 3\\2 & 4 & 0\\-8 & 2 & 1\end{bmatrix}$

13. Show that Helmert’s $2 \times 2$  matrix is its own inverse:  $\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix}$

14. Show that Helmert’s  $3 \times 3$ matrix is its own inverse: $\begin{bmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}}\end{bmatrix}$ .

15. Non-square matrices sometimes have left inverses, where $A^{-1} \cdot A=I$ , or right inverses, where $A \cdot A^{-1}=I$ .  Why can’t non-square matrices have “regular” inverses?

### Vocabulary Language: English

determinant

determinant

The determinant is a single number descriptor of a square matrix. The determinant is computed from the entries of the matrix, and has many properties and interpretations explored in linear algebra.
identity matrix

identity matrix

An identity matrix is a matrix with zeros everywhere except along the diagonal where there are ones.
invertible

invertible

A square matrix is invertible if it has an inverse. If the determinant of a square matrix is zero then the matrix is not invertible.
linearly independent

linearly independent

Three equations are linearly independent if each equation cannot be produced by a linear combination of the other two. Systems that are linearly independent will have just one solution.
Square matrices

Square matrices

Square matrices are matrices in which the number of rows equals the number of columns.
Square matrix

Square matrix

A square matrix is a matrix in which the number of rows equals the number of columns.