# 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)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 $(AA^{-1})_{22}$:

- 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.**

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

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