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.
\(X=A^{-1}B=\begin{pmatrix}9&-4\\-2&1\end{pmatrix} |
\begin{pmatrix}3\\5\end{pmatrix}=\begin{pmatrix}7\\-1\end{pmatrix}\)