Inverse Matrix
The Mathematical Undo Button

Have you ever made a mistake, missed your turn, or gone too far? Sometimes in life, we need a redo, a U-turn, or a chance to undo something.

Well, in mathematics, this is called the inverse. Addition undoes subtraction, multiplication undoes division, square root undoes a square, etc. And an inverse matrix undoes a matrix!

The Inverse Matrix, denoted \(A^{-1}\), allows us to solve and manipulate matrix equations and is similar to how we use the reciprocal and the multiplicative inverse of a nonzero number.

Why We Need Inverse Matrices

But why do we care? What’s the point?

Because we cannot divide by a matrix! Yes, you read that correctly. There is no division of matrices!

Therefore, to manipulate a matrix equation where division is not allowed, the only tool we can use is the inverse.

For example, \(4 \cdot \frac{1}{4}=1\) and \(\frac{1}{4} \cdot 4=1\), which shows how the reciprocal, or inverse of a nonzero number equals 1.

Therefore, we can use this property to our advantage when solving matrix equations because if A is a square \(n \times n\) matrix, then \(A A^{-1}=I\) and \(A^{-1} A=I\), where \(I\) is the identity matrix (i.e., \(\left.a \cdot \frac{1}{a}=1\right)\)

But before we solve matrix equations using inverses, let’s find out how to find an inverse of a matrix.

Finding the Inverse of a 2×2 Matrix

If \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) then \(A^{-1}=\frac{1}{\operatorname{det}(A)}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\), where \(\operatorname{det}(A)=a d-b c\) and \(\operatorname{det}(A) \neq 0\)

Notice that we are dividing by the determinant of \(\mathrm{A}\); thus, if the determinant equals zero (i.e., then the matrix is not invertible.

Invertible and Singular Matrices

Alright, so A is an invertible matrix if it has an inverse and is sometimes called nonsingular. This means that \(\mathrm{A}\) and \(A^{-1}\) are like buddies, as one undoes the other.

However, not all matrices have inverses. Any matrix that is not invertible is called a singular matrix.

Why do we call a matrix that doesn’t have an inverse singular? As a mathematician, I have to say that it is because the determinant of the matrix is zero. But the “eccentric” side of me wants to believe it’s because the word “singular” means exceptional, rare, or even a bit odd. Therefore, if a matrix doesn’t have an inverse, it stands alone as being “unique,” as it doesn’t have a buddy to undo any mistakes – it stands alone.

Example: Identifying a Singular Matrix

Okay, so let’s practice.

Find the inverse of \(A=\left[\begin{array}{ll}3 & 2 \\ 7 & 4\end{array}\right]\)

First, we must calculate the determinant to tell us if the matrix is invertible.

\begin{aligned}
\operatorname{det}(A) & =3(4)-2(7) \\
& =12-14 \\
& =-2
\end{aligned}

Because our determinant is a number other than zero, we know that matrix \(\mathrm{A}\) is invertible (nonsingular). Therefore, we can apply our inverse formula for a \(2 \times 2\) matrix.

\begin{align*}
A^{-1} & =\frac{1}{\operatorname{det}(A)}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
& \Rightarrow A^{-1}=\frac{1}{-2}\left[\begin{array}{cc}
4 & -2 \\
-7 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
-2 & 1 \\
\frac{7}{2} & -\frac{3}{2}
\end{array}\right]
\end{align*}

Easy, right?

Example

Let’s look at another problem.

Find the inverse of \(C=\left[\begin{array}{cc}2 & 4 \\ -3 & -6\end{array}\right]\)

First, we must calculate the determinant to tell us if the matrix is invertible.

\begin{aligned}
\operatorname{det}(C) & =2(-6)-4(-3) \\
& =-12+12 \\
& =0
\end{aligned}

Uh, oh! Well, because the determine is zero, matrix \(\mathrm{C}\) is singular and has no inverse.

Okay, so now it’s time to look at how we can use the Inverse of a Matrix to solve a system of linear equations, as

\begin{align*}
A \vec{x}=\vec{b} \quad \Leftrightarrow \quad \vec{x}=A^{-1} \vec{b}
\end{align*}

Solving Linear Equations with Inverse Matrices

So, all we have to do is find the inverse of \(\mathrm{A}\) and then multiply it by the column vector \(\mathrm{b}\)!

Example: Using Inverse Matrices to Solve a System of Linear Equations

For instance, let’s use the inverse to solve the following system.

\begin{align*}
\left\{\begin{array}{c}
8 x_{1}+5 x_{2}=-9 \\
-7 x_{1}-5 x_{2}=11
\end{array}\right.
\end{align*}

First, let’s write the system of linear equations as a matrix equation \(A \vec{x}=\vec{b}\)

\begin{align*}
\left[\begin{array}{cc}
8 & 5 \\
-7 & -5
\end{array}\right]\left[\begin{array}{l}
x_{1} \\
x_{2}
\end{array}\right]=\left[\begin{array}{c}
-9 \\
11
\end{array}\right]
\end{align*}

Next, we will find the determinant of matrix \(A\)

\begin{aligned}
\operatorname{det}(A) & =8(-5)-5(-7) \\
& =-40+35 \\
& =-5
\end{aligned}

Now we will apply our inverse formula for a \(2 \times 2\) matrix.

\begin{align*}
A^{-1} & =\frac{1}{\operatorname{det}(A)}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
& \Rightarrow A^{-1}=\frac{1}{-5}\left[\begin{array}{cc}
-5 & -5 \\
7 & 8
\end{array}\right]
\end{align*}

Finally, we will find the product of \(\vec{x}=A^{-1} \vec{b}\) by first multiplying the matrices together and then distributing the scalar.

\begin{align*}
\vec{x} & =A^{-1} \vec{b} \\
& =\frac{1}{-5}\left[\begin{array}{cc}
-5 & -5 \\
7 & 8
\end{array}\right]\left[\begin{array}{c}
-9 \\
11
\end{array}\right] \\
& =\frac{1}{-5}\left[\begin{array}{c}
(-5)(-9)+(-5)(11) \\
(7)(-9)+(8)(11)
\end{array}\right] \\
& =\frac{1}{-5}\left[\begin{array}{c}
-10 \\
25
\end{array}\right] \\
& =\left[\begin{array}{c}
2 \\
-5
\end{array}\right]
\end{align*}

So, this means that \(x_{1}=2\) and \(x_{2}=-5\)

Great!

Important Warnings: Invertibility and Order

Yet, there are two critical warnings that you must keep in mind: not every matrix is invertible, and order matters!

We’ve already talked about nonsingular (invertible) and singular (not invertible), but what’s so crucial about order?

As we learned about multiplying matrices in our previous lesson dealing with Matrix Operations, the product of two matrices AB doesn’t always equal BA.

Therefore, we must pay close attention to the fact that when solving matrix equations, the inverse of A must come before our \(b\) vector.

In our video lesson, we will work through several examples of how to move matrices from one side to the other, where we will pay close attention to order and placement.

As you can see, using inverses to solve matrices is pretty straightforward, but what do we do when the matrix is not \(2 \times 2\)?

Algorithm for Finding Inverses of Larger Matrices

Well, we will have to use the inverse algorithm.

The inverse algorithm states that a square \(n \times n\) matrix \(\mathrm{A}\) is invertible if and only if \(\mathrm{A}\) is row equivalent to \(I_{n}\), and in this case, any sequence of elementary row operations that reduces \(\mathrm{A}\) to \(I_{n}\) also transforms \(I_{n}\) to \(A^{-1}\).

In other words, we will augment matrix A with the Identity matrix of the same size.

Then we will use our row echelon techniques to transform the matrix so that:

\begin{equation}
\left[\begin{array}{ll}
A & I
\end{array}\right] \sim\left[\begin{array}{ll}
I & A^{-1}
\end{array}\right]
\end{equation}

Example: Applying the Inverse Algorithm to a 3×3 Matrix

For instance, let’s use the algorithm to find \(A^{-1}\) if \(A=\left[\begin{array}{ccc}1 & 0 & -2 \\ -3 & 1 & 4 \\ 2 & -3 & 4\end{array}\right]\)

First, we augment A with the Identity matrix of the same size.

\begin{align*}
\left[\begin{array}{cccccc}
1 & 0 & -2 & 1 & 0 & 0 \\
-3 & 1 & 4 & 0 & 1 & 0 \\
2 & -3 & 4 & 0 & 0 & 1
\end{array}\right]
\end{align*}

Now, we use Gauss-Jordan Elimination \((\mathrm{RREF})\) so that \(\left[\begin{array}{cc}A & I\end{array}\right] \sim\left[\begin{array}{ll}I & A^{-1}\end{array}\right]\)

\begin{align*}
\left[\begin{array}{cccccc}
1 & 0 & -2 & 1 & 0 & 0 \\
-3 & 1 & 4 & 0 & 1 & 0 \\
2 & -3 & 4 & 0 & 0 & 1
\end{array}\right] & \underset{R R E F}{\sim}\left[\begin{array}{cccccc}
1 & 0 & 0 & 8 & 3 & 1 \\
0 & 1 & 0 & 10 & 4 & 1 \\
0 & 0 & 1 & 7 / 2 & 3 / 2 & 1 / 2
\end{array}\right]
\end{align*}

Notice that the Identity matrix moved from back to front!

This means:

\begin{equation}
A^{-1}=\left[\begin{array}{ccc}
8 & 3 & 1 \\
10 & 4 & 1 \\
7 / 2 & 3 / 2 & 1 / 2
\end{array}\right]
\end{equation}

Next Steps

In this video, you will:

• Work through the process together
• Examine the definition of Elementary Matrices and how they differ from the Identity matrix by one single elementary row operation
• Understand the importance of Elementary Matrices for future lessons on the invertible matrix theorem and properties of determinants
• Appreciate the significance of inverses for the journey ahead

Get ready to jump right in and explore!

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