 # Determinant of a 2 x 2 Matrix – Explanation & Examples

The determinant of a matrix is a scalar value that is quite important in linear algebra. We can solve the linear system of equations with the determinant and find the inverse of square matrices. The simplest determinant is that of a $2 \times 2$ matrix.

The determinant of a 2 x 2 matrix is a scalar value that we get from subtracting the product of top-right and bottom-left entry from the product of top-left and bottom-right entry.

In this lesson, we will look at the formula for a $2 \times 2$ matrix and find the determinant of a $2 \times 2$ matrix. Several examples will help us engulf the information thoroughly. Let us start!

## What is the Determinant of a Matrix?

Recall that a matrix’s determinant is a scalar value that results from certain operations done on the matrix. We can denote the determinant of a matrix in $3$ ways:

Consider the $2 \times 2$ matrix shown below:

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

We can denote its determinant in the following $3$ ways: For the $2 \times 2$ matrix A, we denote its determinant by writing $det(A)$, $| A |$, or $A = \begin{vmatrix} { a } & { b } \\ { c } & { d } \end {vmatrix}$.

## How to find the Determinant of a 2 x 2 Matrix

First of all, we can only calculate the determinant for square matrices! There aren’t any determinants for non-square matrices.

There is a formula (specifically, an algorithm) to find the determinant of any square matrices. But that is out of the scope of this lesson, and we won’t be looking at it here. We will be checking out the determinant of the simplest square matrix, the $2 \times 2$ matrix.

Below, we look at the formula for the determinant of a $2 \times 2$ matrix and show several examples of finding the determinant of a $2 \times 2$ matrix.

## Determinant of a 2 x 2 Matrix Formula

Consider the $2 \times 2$ matrix shown below:

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

The formula for the determinant of a $2 \times 2$ matrix is shown below:

$det( A ) = | A | = \begin{vmatrix} { a } & { b } \\ { c } & { d } \end {vmatrix} = ad – bc$

Note: We used $3$ different notations to show the determinant of this matrix.

The determinant of a 2 x 2 matrix is a scalar value that we get from subtracting the product of top-right and bottom-left entry from the product of top-left and bottom-right entry. Let’s calculate the determinant of Matrix $B$ shown below:

$B = \begin{bmatrix} { 0 } & { 4 } \\ { – 1 } & { 10 } \end {bmatrix}$

Using the formula just learned, we can find the determinant:

$det( B ) = | B | = \begin{vmatrix} { 0 } & { 4 } \\ { – 1 } & { 10 } \end {vmatrix}$

$= ( 0 ) ( 10 ) – ( 4 ) ( – 1 )$

$= 0 + 4$

$= 4$

The determinant of matrix $B$ is calculated to be $4$.

Be careful with signs! Since there is a minus sign in between the terms $ad$ and $bc$ in the determinant of a $2 \times 2$ matrix formula, it is easy to get arithmetic errors when the elements of the matrix contain negative numbers!

We will look at several examples to enhance our understanding further.

#### Example 1

Given $D = \begin{bmatrix} { – 3 } & { 1 } \\ { 6 } & { – 4 } \end {bmatrix}$, find $| D |$.

Solution

We have to find the determinant of the $2 \times 2$ matrix $D$ shown above. Let’s use the formula and find the determinant.

Shown below:

$det( D ) = | D | = \begin{vmatrix} { – 3 } & { 1 } \\ { 6 } & { – 4 } \end {vmatrix}$

$= ( – 3 ) ( – 4 ) – ( 1 ) ( 6 )$

$= 12 – 6$

$= 6$

The determinant of Matrix $D$ is $6$.

#### Example 2

Given $A = \begin{bmatrix} { – 14 } & { – 2 } \\ { – 6 } & { – 3 } \end {bmatrix}$, find $| A |$.

Solution

Matrix $A$ is a $2 \times 2$ square matrix. To find its determinant, we use the formula, making sure to be extra careful with signs! The process is shown below:

$det( A) = | A | = \begin{vmatrix} { – 14 } & { – 2 } \\ { – 6 } & { – 3 } \end {vmatrix}$

$= ( – 14 ) ( – 3 ) – ( – 2 ) ( – 6 )$

$= 42 – 12$

$= 30$

The determinant of Matrix $A$ is $30$.

#### Example 3

Calculate the determinant of Matrix $K$ shown below:

$K = \begin{bmatrix} { 8 } & { 24 } \\ { – 4 } & { – 12 } \end {bmatrix}$

Solution

We will use the formula for the determinant of a $2 \times 2$ matrix to calculate the determinant of Matrix $K$. Shown below:

$det( K ) = | K | = \begin{vmatrix} { 8 } & { 24 } \\ { – 4 } & { – 12 } \end {vmatrix}$

$= ( 8 ) ( – 12 ) – ( 24 ) ( – 4 )$

$= – 96 – ( – 96 )$

$= – 96 + 96$

$= 0$

The determinant of this matrix is $0$!

This is a special type of matrix. It is a non-invertible matrix and is known as a singular matrix. Check this article out to know more about singular matrices!

#### Example 4

Find $m$ given $\begin{vmatrix} { – 3 } & { 4 } \\ { m } & { – 12 } \end {vmatrix} = – 36$.

Solution

In this problem, we are already given the determinant and have to find an element of the matrix, $m$. Let’s plug it into the formula and do some algebra to figure out $m$. The process is shown below:

$\begin{vmatrix} { – 3 } & { 4 } \\ { m } & { – 12 } \end {vmatrix} = – 36$

$( – 3 ) ( – 12) – ( 4 ) ( m ) = – 36$

$36 – 4m = – 36$

$4m = 36 + 36$

$4 m = 72$

$m = \frac{ 72 }{ 4 }$

$m = 18$

The value of m is $18$.

Now, it’s your turn to practice some questions!

### Practice Questions

1. Find the determinant of the matrix shown below:
$B = \begin{bmatrix} { – \frac{ 1 }{ 2 } } & { – \frac{ 1 }{ 6 } } \\ { – 10 } & { 12 } \end {bmatrix}$

2. Find $t$ given $\begin{vmatrix} { 8 } & { t } \\ { – 2 } & { \frac{ 1 }{ 4 } } \end {vmatrix} = 42$.

3. Consider matrices $A$ and $B$ shown below:
$A = \begin{bmatrix} { 2 } & { – 3 } \\ { x } & { – 8 } \end {bmatrix}$
$B = \begin{bmatrix} { x } & { 12} \\ { – 2 } & { – 5 } \end {bmatrix}$
If the determinant of both the matrices are equal ($| A | = | B |$), find out the value of $x$.

1. Matrix $B$ is a $2 \times 2$ square matrix. Let’s find the determinant by using the formula we learned in this lesson. Some of the elements of Matrix $B$ are fractions. It will make the calculation a tiny bit more tedious. Otherwise, everything else is the same

The process of finding the determinant is shown below:

$det( B ) = | B | = \begin{vmatrix} { – \frac{ 1 }{ 2 } } & { – \frac{ 1 }{ 6 } } \\ { – 10 } & { 12 } \end {vmatrix}$

$= ( – \frac{ 1 }{ 2 } ) ( 12 ) – ( – \frac{ 1 }{ 6 } ) ( – 10 )$

$= – 6 – \frac{ 5 }{ 3 }$

$= -6\frac{ 5 }{ 3 }$

Thus, $| B | = -6\frac{ 5 }{ 3 }$.

2. In this problem, we are already given the determinant and have to find an element of the matrix, $t$. Let’s plug it into the formula and do some algebra to figure out $t$. The process is shown below:

$\begin{vmatrix} { 8 } & { t } \\ { – 2 } & { \frac{ 1 }{ 4 } } \end {vmatrix} = 42$

$( 8 ) ( \frac{ 1 }{ 4 } ) – ( t ) ( – 2 ) = 42$

$2 + 2t = 42$

$2t = 42 – 2$

$2t = 40$

$t = \frac{ 40 }{ 2 }$

$t = 20$

The value of t is $20$.

3. Using the formula for the determinant of a $2 \times 2$ matrix, we can write the expressions for the determinant of Matrix $A$ and Matrix $B$.

Determinant of Matrix $A$:
$| A | = \begin{vmatrix} { 2 } & { – 3 } \\ { x } & { – 8 } \end {vmatrix}$
$| A | = ( 2 )( – 8 ) – ( – 3 )( x )$
$| A | = – 16 + 3x$

Determinant of Matrix $B$:
$| B | = \begin{vmatrix} { x } & { 12} \\ { – 2 } & { – 5 } \end {vmatrix}$
$| B | = ( x )( – 5 ) – ( 12 )( – 2 )$
$| B | = – 5x + 24$

Since both the determinants are equal, we equate both expressions and solve for $x$. The algebraic process is shown below:

$| A | = | B |$

$– 16 + 3x = – 5x + 24$

$3x + 5x = 24 + 16$

$8x = 40$

$x = \frac{ 40 }{ 8 }$

$x = 5$

The value of $x$ is $5$.