Evaluating 2 x 2 Determinants And Matrix Inverses

# Determinants for 2 x 2 Matrices

We will now begin to look at an important value associated to every square matrix known as the determinant of that matrix. We will start by looking at determinants for $2 \times 2$ matrices which conveniently have an easy formula for computational purposes. We will subsequently look at computing determinants for larger square matrices.

 Definition: Given a $2 \times 2$ matrix $A$ in the form $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$, the Determinant of the $2 \times 2$ Matrix $A$ denoted $\det (A) = ad - bc$. Note that this definition applies only towards determinants of matrices that have size $2 \times 2$.

For example, consider the following $2 \times 2$ matrix:

(1)
\begin{align} A = \begin{bmatrix} 5 & 2\\ 1 & 3 \end{bmatrix} \end{align}

We note that $a = 5$, $b = 2$, $c = 1$, and $d = 3$. Therefore using the formula in the definition, we get that $\det (A) = ad - bc = (5)(3) - (2)(1) = 13$.

We will now look at an important theorem where we can use the determinant of a $2 \times 2$ matrix in order to quickly test if the matrix has an inverse and calculate it.

 Theorem 1: A $2 \times 2$ matrix $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$ is invertible if $\det(A) = ad - bc ≠ 0$. The inverse of $A$ can be obtained with the following formula $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} = \begin{bmatrix} \frac{d}{ad - bc} & - \frac{b}{ad - bc}\\ - \frac{c}{ad - bc} & \frac{a}{ad - bc} \end{bmatrix}$. If $ad - bc = 0$, then $A$ is not invertible.
• Proof of Theorem 1: Assume that $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \frac{d}{ad - bc} & - \frac{b}{ad - bc}\\ - \frac{c}{ad - bc} & \frac{a}{ad - bc} \end{bmatrix}$. We will proceed to show that $AA^{-1} = I_2$. We will first calculate all of the value of all entries in the product $AA^{-1}$, that is $(AA^{-1})_{11}$, $(AA^{-1})_{12}$, $(AA^{-1})_{21}$, and $(AA^{-1})_{22}$:
(2)
(3)
\begin{align} (AA^{-1})_{12} = \frac{-ab}{ad - bc} + a \cdot \frac{b}{ad - bc} \\ (AA^{-1})_{12} = \frac{-ab + ab}{ad - bc} = 0 \end{align}
(4)
\begin{align} (AA^{-1})_{21} = \frac{cd}{ad - bc} - \frac{cd}{ad - bc} \\ (AA^{-1})_{21} = \frac{cd - cd}{ad - bc} = 0 \end{align}
(5)
• When $A$ and $A^{-1}$ are multiplied through we obtain $I_2 = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$. We also note that if a matrix $A$ is invertible, there exists only one unique inverse which we have found. We note that if $ad - bc = 0$, then $\frac{1}{ad-bc} = \frac{1}{0}$, which is undefined. $\blacksquare$

## Example 1

Given the following matrix, evaluate the inverse by calculating its determinant.

(6)
\begin{align} A = \begin{bmatrix} 4 & -2\\ 3 & 6 \end{bmatrix} \end{align}

We first calculate that $\det (A) = ad - bc = (4)(6) - (-2)(3) = 30$. By the formula in theorem 1 then:

(7)
\begin{align} \quad A^{-1} = \frac{1}{30} \begin{bmatrix} 6 & 2\\ -3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{6}{30} & \frac{2}{30}\\ -\frac{3}{30} & \frac{4}{30} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{1}{15}\\ -\frac{1}{10} & \frac{2}{15} \end{bmatrix} \end{align}

You can verify that this inverse is correct by $AA^{-1} = I_2$.