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What Is a Determinant? · Linear Algebra | Math Infinitum

Practice

Lesson goal

understand what a determinant is;
recognize which matrices have determinants;
compute the determinant of a $1\times1$ matrix;
compute the determinant of a $2\times2$ matrix;
avoid the most common sign mistake;
use the determinant formula in simple symbolic examples.

A Matrix Can Produce a Number

A matrix is an array of numbers.

But for a square matrix, we can associate a single number with the entire matrix. Consider

A=\begin{pmatrix}2&1\\3&4\end{pmatrix}.

From these four entries we can compute one number:

2\cdot4-1\cdot3=5.

This number is called the determinant of the matrix. We write

\det A=5.

What Is a Determinant?

If $A$ is a square matrix, its determinant can be written as $\det A$ or using vertical bars: $|A|$ .

For example,

\begin{pmatrix}1&2\\3&4\end{pmatrix}

is square, so we can calculate its determinant. But

\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}

is not square.

Try it

Which matrix can have a determinant?

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The Simplest Determinant

For a matrix of order $1$ , $A=(a)$ , the determinant is simply the only entry of the matrix.

\boxed{\det A=a}

For example, $A=(7)$ . Then

\det A=7.

Try it

Let $A=(-4)$ . What is $\det A$ ?

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The 2×2 Formula

Consider a general $2\times2$ matrix

A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}.

Its determinant is

\boxed{\det A=a_{11}a_{22}-a_{12}a_{21}}

or, using determinant notation,

\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}.

Multiply Along the Diagonals

For

\begin{vmatrix}a&b\\c&d\end{vmatrix},

first multiply along the main diagonal: $a\cdot d$ . Then multiply along the other diagonal: $b\cdot c$ . Finally subtract:

\boxed{ad-bc}.

Example

Find the determinant:

\begin{vmatrix}2&-3\\5&6\end{vmatrix}.

Solution. Using

\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,

we get

\det A=2\cdot6-(-3)\cdot5.

Therefore,

=12-(-15)

and hence

\boxed{\det A=27}.

Watch the Minus Sign

The formula is $ad-bc$ . Suppose

A=\begin{pmatrix}2&-3\\5&6\end{pmatrix}.

Then

bc=(-3)\cdot5=-15.

ad-bc=12-(-15).

Subtracting a negative number means adding:

12+15=27.

Try it

Compute $\begin{vmatrix}3&2\\1&4\end{vmatrix}$ .

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Try it

Compute $\begin{vmatrix}5&-2\\3&1\end{vmatrix}$ .

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Try it

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , the determinant is $ad+bc$ .

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Try it

For the matrix $A$ , select the two entries whose product forms the first term of the determinant.

$A=\begin{pmatrix}4 & 7 \\ 2 & 5\end{pmatrix}$

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Try it

Compute $\det\begin{pmatrix}4&7\\2&5\end{pmatrix}$ .

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Determinants Can Contain Variables

The entries of a matrix do not have to be ordinary numbers. For example,

A=\begin{pmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{pmatrix}.

Using the $2\times2$ determinant formula,

\det A=\cos\alpha\cos\alpha-\sin\alpha(-\sin\alpha).

Therefore,

\det A=\cos^2\alpha+\sin^2\alpha.

Using

\cos^2\alpha+\sin^2\alpha=1,

we obtain

\boxed{\det A=1}.

Try it

Consider $A=\begin{pmatrix}x&0\\0&y\end{pmatrix}$ . What is $\det A$ ?

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A Useful Pattern

For a diagonal $2\times2$ matrix

D=\begin{pmatrix}a&0\\0&d\end{pmatrix},

the determinant is

\det D=ad-0.

Therefore,

\boxed{\det D=ad}.

So in this case, the determinant is simply the product of the diagonal entries.

Try it

Compute $\det\begin{pmatrix}-3&0\\0&5\end{pmatrix}$ .

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A Determinant Can Be Zero

Consider

A=\begin{pmatrix}1&2\\2&4\end{pmatrix}.

Then

\det A=1\cdot4-2\cdot2

\det A=4-4=\boxed{0}.

There is nothing unusual about a determinant being zero. Later, we will learn important properties that explain when this happens.

Try it

Which matrix has determinant $0$ ?

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Try it

Find $x$ if $\begin{vmatrix}x&2\\3&4\end{vmatrix}=10$ .

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Do Not Add the Products

For

\begin{vmatrix}a&b\\c&d\end{vmatrix},

the formula is

\boxed{ad-bc}.

Not $ad+bc$ .
Not $ac-bd$ .
And not $ab-cd$ .

The positions of the entries matter.

What You Learned

A determinant is a number associated with a square matrix. It can be written as $\det A$ or $|A|$ .

For a $1\times1$ matrix, $A=(a)$ , we have

\boxed{\det A=a}.

For a $2\times2$ matrix,

A=\begin{pmatrix}a&b\\c&d\end{pmatrix},

we have

\boxed{\det A=ad-bc}.

A useful memory rule is:

\boxed{\begin{aligned}&\text{main diagonal product}\\-\;&\text{other diagonal product}\end{aligned}}

Final checkpoint

Try it

Compute $\begin{vmatrix}2&3\\4&5\end{vmatrix}$ .

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Try it

Compute $\begin{vmatrix}-1&4\\2&3\end{vmatrix}$ .

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Try it

Only square matrices are considered when defining determinants in this lesson.

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Try it

Let $A=\begin{pmatrix}x&1\\2&3\end{pmatrix}$ . If $\det A=7$ , what is $x$ ?

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Try it

Which formula correctly gives $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}$ ?

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Completion

Lesson complete

Great work! You now know how to:

recognize a determinant;
calculate determinants of order 1;
calculate determinants of order 2;
use the diagonal method for $2\times2$ determinants;
work with negative and symbolic entries.