This page is intended to be a part of the Numerical Analysis section of Math Online. Similar topics can also be found in the Linear Algebra section of the site.

The Invertibility of a Matrix

Recall from the Matrix Arithmetic page that a square $n \times n$ matrix $A$ is said to be invertible if there exists an $n \times n$ matrix $B$ such that $AB = I_n = BA$ and we usually denote the inverse of $A$ as $B = A^{-1}$. Furthermore, if no such matrix $B$ exists, then $A$ is said to be singular or noninvertible.

We will now briefly discuss the invertibility of $2 \times 2$ matrices and their importance when solving systems of linear equations. We will first start with the most elementary case of square matrices - $2 \times 2$ matrices. Consider the following arbitrary $2 \times 2$ square matrix:

(1)

\begin{align} \quad A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \end{align}

The following theorem gives us a way to find the inverse matrix $A^{-1}$ provided that it exists.

Theorem 1: Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then the inverse of $A$ can be obtained with the formula $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} \frac{d}{ad - bc} & -\frac{b}{ad - bc} \\ -\frac{c}{ad - bc} & \frac{a}{ad - bc} \end{bmatrix}$. Furthermore, $A$ is invertible if and only if $ad - bc \neq 0$.

• Proof: We can carry this proof out by multiplying $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with the suggested formula for $A^{-1}$, call it $B = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. Let's compute the product, $AB$ first. Doing this entry-wise and we have that the first entry in the first row of this product is:

(2)

\begin{align} \quad \frac{ad - bc}{ad - bc} = 1 \end{align}

• The second entry in the first row of this product is:

(3)

\begin{align} \quad \frac{-ab + ab}{ad - bc} = 0 \end{align}

• The first entry in the second row of this product is:

(4)

\begin{align} \quad \frac{cd - cd}{ad - bc} = 0 \end{align}

• The second entry in the second row of this product is:

(5)

\begin{align} \quad \frac{-bc + ad}{ad - bc} = 1 \end{align}

• Therefore $AB = I$. We now only need to show that $BA = I$, which can be shown in a similar manner.

• Now if $A$ is invertible the only one inverse exists and it must be in the form described above, in which case $ad - bc \neq 0$ otherwise $A^{-1}$ would not be well defined. Conversely, if $ad - bc \neq 0$, then $A^{-1}$ is well defined and so $A$ is invertible. $\blacksquare$

For example, consider the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Note that $(1)(4) - (2)(3) = -2 \neq 0$, and so Theorem 1 guarantees that $A^{-1}$ exists and $A^{-1}$ is given by:

(6)

\begin{align} \quad A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \end{align}

You should verify that $AA^{-1} = I = A^{-1}A$ as an exercise.

Now we will soon see that a linear system of $m$ equations in $n$ unknowns can be nicely represented using matrices. If we are lucky enough to have a linear system of $n$ equations in $n$ unknowns, then the existence of a unique solution can be guaranteed provided that the matrix that represents the system is invertible. Don't worry too much about this at the present time if you're not familiar with this concept. We will delve deeper into this later. For now, let's consider an example to get us motived. Consider the following system of of $2$ equations in $2$ unknowns:

(7)

\begin{align} \quad x_1 + 2x_2 = 6 \\ \quad 3x_1 + 4x_2 = -3 \end{align}

Suppose that we create a matrix (known as an augmented matrix) of the coefficients of the system. Then we get the matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ from earlier. Furthermore, let $x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ and let $b = \begin{bmatrix} 6 \\ -3 \end{bmatrix}$. Then the system above can be represented in the form $Ax = b$, that is:

(8)

\begin{align} \quad Ax = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \end{bmatrix} \end{align}

Now we already know that $A^{-1}$ exists as we calculated it earlier. So for the equation $Ax = b$, if we multiply both sides by $A^{-1}$ we get that $A^{-1}Ax = A^{-1}b$, or equivalently, $x = A^{-1}b$. Thus the solution to this system of equations is obtained by taking the matrix $A^{-1}$ and multiplying it by the matrix $b$. In doing so we get that:

(9)

\begin{align} \quad A^{-1} b = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 6 \\ -3 \end{bmatrix} = \begin{bmatrix} -15 \\ \frac{21}{2} \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \end{align}

So the solution to our system of equations is $(x_1, x_2) = \left ( -15, \frac{21}{2} \right )$. You should verify that this is indeed the point of intersects of the lines $x_1 + 2x_2 = 6$ and $3x_1 + 4y_1 = -3$.

So, determining the inverse of a $2 \times 2$ square matrix is relatively easy and useful for solving systems of two equations with two unknowns, and it's not hard to see that if $A$ is an $n \times n$ matrix, then we can obtain the solution corresponding to the system represented by $A$ as $x = A^{-1} b$ provided that $A$ is invertible. Of course, the next question is to what sort of formulas/methods we can use in order to determine the inverse or invertibility of a larger square matrix. Such formulas exists for determining the inverse of a $3 \times 3$ matrix and higher, however, the formulas become tediously long and not that useful. Instead, there are other techniques to determine the invertibility of a matrix without necessarily solving for the inverse as the process can become cumbersome for matrices with thousands of entries.

Before we end off this page, we'll note that if $A$ represents a system of linear equations and is invertible, then $x = A^{-1}b$ gives us a unique solution to the system. Suppose that $x = A^{-1}b$ and $x = B^{-1}b$. Then $A^{-1}b = B^{-1}b$ and multiplying both sides of the equation above on the left by $A$ and we have that $b = AB^{-1} b$. So $AB^{-1} = I$ and so $B^{-1} = A^{-1}$, and thus our solution is unique. Furthermore, if we have the equation $Ax = 0$ and $A$ is invertible, then $x = 0$, and so the unique solution that for which all variables are equal to $0$. Such a solution is known as a Trivial Solution and such a system is known as a Homogenous System of Linear Equations.