# Inverses of Matrices in Algebra

These are my notes on inverses of matrices.

**Matrix Inverses**

If you have the matrix:

\[A=\begin{pmatrix}1&0&h\\0&1&k\\0&0&1\end{pmatrix}\]

It is used to translate a point (x,y) horizontally h units and vertically k

units. The translation is to the right if h>0 and to the left if k<0. Similarly,

the translation is upward if k>0 and downward if k<0. A point (x,y) is

represented by the 3*1 column matrix:

\[X=\begin{pmatrix}x\\y\\1\end{pmatrix}\]

The third element in X is always equal to 1. For example, the point (-1,2)could

be translated 3 units right and 4 units downward by computing the matrix

product.

\[AX=\begin{pmatrix}1&0&3\\0&1&-4\\0&0&1\end{pmatrix}*\begin{pmatrix}-1&2&1\end{pmatrix}=Y\]

Its new location is (2,-2). In the matrix A, h=3 and k=-4. If A translates a

point 3 units right and 4 units downward, then the inverse matrix translates a

point 3 units left and 4 units upward. This would return a point to its original

position after being translated by A. Therefore, the inverse matrix of A,

denoted \(A^{-1}\), is given by:

\[A^{-1}=\begin{pmatrix}1&0&-3\\0&1&4\\0&0&1\end{pmatrix}\]

The n*n Identity Matrix

Denoted \(I_{n}\), has only 1's on its main diagonal and 0's elsewhere. Some

examples of identity matrices are:

\[\begin{pmatrix}1&0\\0&1\end{pmatrix}\]

and

\[\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\]**Inverse of a Square Matrix**

If \(A^{-1}\) exists, then A is invertible or nonsingular. On the other hand, if

a matrix A is not invertible, then it is singular. Not every matrix has an

ivnerse. For example, the zero matrix with dimension 3*3 is given by:

\[\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\]

This matrix does not have an inverse. the product of \(O_{3}\) with any 3*3

matrix B would again be \(O{3}\), rather than the identity matrix \(I_{3}\).**Example 1**

Verifying an Inverse

Determine if B is the inverse of A where:

\[A=\begin{pmatrix}5&3\\-3&-2\end{pmatrix}\text { and }

B=\begin{pmatrix}2&3\\-3&-5\end{pmatrix}\]

Solution:

For B to be the inverse of A, it must satisfy the equations \(AB=I_{2}\) and

\(BA=I_{2}\)

AB=\[\begin{pmatrix}5&3\\-3&-2\end{pmatrix}*\begin{pmatrix}2&3\\-3&-5\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\]

BA=\[\begin{pmatrix}2&3\\-3&-5\end{pmatrix}*\begin{pmatrix}5&3\\-3&-2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\]

B is the inverse of A.**Example 2**

Interpreting an Inverse Matrix

the matrix A can be used to rotate a point 90 degrees clockwise about the origin

where:

\[A=\begin{pmatrix}0&1&0\\0&0&1\end{pmatrix} \text{ and }

A^{-1}=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}\]

Use A to rotate the point (-2,0) clockwise 90 degrees about the origin.

Make a conjecture about the effect of \(A^{-1}\) on the resulting point.

Test this conjecture.

Solution:

First, let the point (-2,0) be represented by the column matrix

\[X=\begin{pmatrix}-2\\0\\1\end{pmatrix}\]

Then compute AX.

\[\begin{pmatrix}0&1&0\\{-1}&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}{-2}\\0\\1\end{pmatrix}=\begin{pmatrix}0\\2\\1\end{pmatrix}=Y\]

If the point (-2,0) is rotated 90 degrees clockwise about the origin, its new

location is (0,2).

Since \(A^{-1}\) represents the inverse operation of A, the inverse will rotate

the point located at (0,2) counterclockwise 90 degrees, back to (-2,0).

this conjecture is correct since:

\[A^{-1}Y=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}0\\2\\1\end{pmatrix}=\begin{pmatrix}-2\\0\\1\end{pmatrix}=X\]**Finding Inverses Symbolically**

The inverse matrix can be found symbolically by first forming the augmented

matrix \(A|I_{n}\), and then performing matrix row operations, until the left

side of the augmented matrix becomes the identity matrix. The resulting

augmented matrix can be written as \(I_{n}|A^{-1}\), where the right side of the

matrix is \(A^{-1}\). **Example 3**

Finding an inverse symbolically.

Find \(A^{-1}\) if:

\[A=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}\]

Solution:

Begin by forming the following 3*6 augmented matrix with the 3*3 identity matrix

on the right half.

\(\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}\) | \(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\)

Next we use row transformations to obtain the 3*3 identity on the left side.

Negate the elements in row 2 and then interchange row 1 and row 2. The same row

transformations are also applied to the right side of the augmented matrix.

\(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\) |

\(\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}\)

Because the left side of the augmented matrix is now the 3*3 identity, we can

stop. The right side of the augmented matrix is \(A^{-1}\):

\[A^{-1}=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}\]

Many times finding inverses requires several steps of row transformations.**Example 4**

Finding the inverse of a 2*2 matrix symbolically

Find \(A^{-1}\) if:

\[A=\begin{pmatrix}1&4\\2&9\end{pmatrix}\]

Solution:

Begin by forming a 2*4 augmented matrix. Perform matrix row operations to obtain

the identity matrix on the left side, and perform the same operation on the

right side of this matrix.

\(\begin{pmatrix}1&4\\2&9\end{pmatrix}\) | \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\)

Yields: \(\begin{pmatrix}1&4\\0&1\end{pmatrix}\) |

\(\begin{pmatrix}1&0\\0&1\end{pmatrix}\)

Now: \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\) |

\(\begin{pmatrix}9&-4\\-2&1\end{pmatrix}\)

Since the 2*2 identity matrix appears on the left side, it follows that the

right side equals \(A^{-1}\):

\[A^{-1}=\begin{pmatrix}9&-4\\-2&1\end{pmatrix}\]**Example 5**

Finding the inverse of a 3*3 matrix symbolically

Find \(A^{-1}\) if:

\[A=\begin{pmatrix}1&0&1\\2&1&3\\_1&1&1\end{pmatrix}\]

Solution:

Begin by forming the following 3*6 augmented matrix. Perform matrix row

operations to obtain the identity matrix on the left side, and perform the same

operation on the right side of this matrix.

\(\begin{pmatrix}1&0&1\\2&1&3\\-1&1&1\end{pmatrix} |

\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\)

\(\begin{pmatrix}1&0&1\\0&1&1\\0&1&2\end{pmatrix} |

\begin{pmatrix}1&0&0\\-2&1&0\\1&0&1\end{pmatrix}\)

\(\begin{pmatrix}1&0&1\\0&1&1\\0&0&1\end{pmatrix} |

\begin{pmatrix}1&0&0\\-2&1&0\\3&-1&1\end{pmatrix}\)

\(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} |

\begin{pmatrix}-2&1&-1\\-5&2&-1\\3&-1&1\end{pmatrix}\)

If it is not possible to obtain the identity matrix on the left side of the

augmented matrix by using row operations, then \(A^{-1}\) does not exist.**Representing Linear Systems with Matrix Equations**

Linear systems were solved using Gaussian elimination with backward

substitution. This method used an augmented matrix to represent a system of

linear equations. A system of linear equations can also be represented by a

matrix equation.

\[3x-2y+4z=5\]

\[2x+y+3z=9\]

\[-x+5y-2z=5\]

Let A,X, and B be matrices defined as:

\[A=\begin{pmatrix}3&-2&4\\2&1&3\\-1&5&-2\end{pmatrix}\]

\[X=\begin{pmatrix}x\\y\\z\end{pmatrix}\]

\[B=\begin{pmatrix}5\\9\\5\end{pmatrix}\]

The matrix product AX is:

\[\begin{pmatrix}3x&-2y&4z\\2x&y&3z\\-x&5y&-2z\end{pmatrix}\]

This gives:

\[\begin{pmatrix}3x&-2y&4z\\2x&y&3z\\-x&5y&-2z\end{pmatrix}\]

The matrix operation AX=B simplifies to:

\[\begin{pmatrix}5\\9\\5\end{pmatrix}\]

This matrix equation AX=B is equivalent to the original system of linear

equations. Any system of linear equations can be represented by a matrix

equation in the form AX=B.**Example 6**

Representing linear systems with matrix equations

Represent the system of linear equations in the form AX=B

\(3x-4y=7\) | \(-x+6y=-3\)

Solution:

This lionear system comprises two equations and two variables. This equivalent

matrix equation is:

\(AX=\begin{pmatrix}3&-4\\-1&6\end{pmatrix} |

\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}7\\-3\end{pmatrix} = B\)**Solving Linear Systems with Inverses**

The matrix equation AX=B can be solved by using \(A^{-1}\) if it exists.

\(AX=B\)

\(A^{-1}AX=A^{-1}B\)

\(I_{n}X=A^{-1}B\)

\(X=A^{-1}B\)

To solve a linear system, multiply each side of the matrix equation AX=B by

\(A^{-1}\), if it exists. The solution to the system is unique and can be

written as \(X=A^{-1}B\). Since matrix multiplication is not commutative, it is

essential to multiple each side of the equation on the left by \(A^{-1}\). That

is, \(X=A^{-1}B \not = BA^{-1}\) in general.**Example 7**

Solving a linear system using the inverse of a 2*2 matrix.

Write the linear system as a matrix equation in the form \(AX=B\). Find

\(A^{-1}\) and solve for X.

\(x+4y=3\)

\(2x+9y=5\)

Solution:

The linear system can be written as:

\(AX=\begin{pmatrix}1&4\\2&9\end{pmatrix} |

\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\5\end{pmatrix}=B\)

The matrix \(A^{-1}\) was already found. So, we can solve for X without having

to find \(A^{-1}\) first.

\(X=A^{-1}B=\begin{pmatrix}9&-4\\-2&1\end{pmatrix} |

\begin{pmatrix}3\\5\end{pmatrix}=\begin{pmatrix}7\\-1\end{pmatrix}\)