<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

Determinants

Number calculated from the entries in a matrix primarily through multiplication and subtraction.

Estimated16 minsto complete
%
Progress
Practice Determinants
Progress
Estimated16 minsto complete
%
Determinant of Matrices

A determinant is a number computed from the entries in a square matrix.  It has many properties and interpretations that you will explore in linear algebra.  This concept is focused on the procedure of calculating determinants.  Once you know how to calculate the determinant of a $2 \times 2$  matrix, then you will be able to calculate the determinant of a $3 \times 3$ matrix.  Once you know how to calculate the determinant of a $3 \times 3$  matrix you can calculate the determinant of a $4 \times 4$  and so on.

A logical question about determinants is where does the procedure come from?  Why are determinants defined in the way that they are?

Guidance

The determinant of a matrix  $A$ is written as $|A|$ .  For a $2 \times 2$  matrix $A$ , the value is calculated as:

$A &= \begin{bmatrix}a & b\\c & d\end{bmatrix}\\\text{det} \ A &= |A|=\begin{vmatrix}a & b\\c & d\end{vmatrix}=ad-bc$

Notice how the diagonals are multiplied and then subtracted.

The determinant of a $3 \times 3$  matrix is more involved.

$B=\begin{bmatrix}a & b & c\\d & e & f\\g & h & i\end{bmatrix}$

Usually you will start by looking at the top row, although any row or column will work.  Then use the checkerboard pattern for signs (shown below) and create smaller $2 \times 2$  matrices.

$\begin{bmatrix}+ & - & +\\- & + & -\\+ & - & +\end{bmatrix}$

The smaller  $2 \times 2$ matrices are the entries that remain when the row and column of the coefficient you are working with are ignored.

$\text{det} \ B=|B|=+a \cdot \begin{vmatrix}e & f\\h & i\end{vmatrix}-b \cdot \begin{vmatrix}d & f\\g & i\end{vmatrix}+c \cdot \begin{vmatrix}d & e\\g & h\end{vmatrix}$

Next take the determinant of the smaller $2 \times 2$  matrices and you get a long string of computations.

$&= +a(ei-fh)-b(di-fg)+c(dh-eg)\\&= aei-afh-bdi+bfg+cdh-ceg\\&= aei+bfg+cdh-ceg-afh-bdi$

Most people do not remember this sequence.  A French mathematician named Sarrus demonstrated a great device to memorize the computation of the determinant for $3 \times 3$  matrices.  The first step is simply to copy the first two columns to the right of the matrix. Then draw three diagonal lines going down and to the right.

$B=\begin{bmatrix}a & b & c\\d & e & f\\g & h & i\end{bmatrix}$

Notice that they correspond exactly to the three positive terms of the determinant demonstrated above.  Next draw three diagonals going up and to the right. These diagonals correspond exactly to the three negative terms.

$\text{det} \ B=aei+bfg+cdh-ceg-afh-bdi$

Sarrus’s rule does not work for the determinants of matrices that are not of order $3 \times 3$ .

Example A

Find $\text{det} \ A$  for $A=\begin{bmatrix}3 & 2\\1 & 5\end{bmatrix}$

Solution:  $\begin{vmatrix}3 & 2\\1 & 5\end{vmatrix}=3 \cdot 5-2 \cdot 1=15-2=13$

Example B

Find $\text{det} \ B$  for $B=\begin{bmatrix}3 & 2 & 1\\5 & 0 & 2\\2 & 1 & 5\end{bmatrix}$

Solution:

$\begin{vmatrix}3 & 2 & 1\\5 & 0 & 2\\2 & 1 & 5\end{vmatrix} &=3 \begin{vmatrix}0 & 2\\1 & 5\end{vmatrix}-2 \begin{vmatrix}5 & 2\\2 & 5\end{vmatrix}+1 \begin{vmatrix}5 & 0\\2 & 1\end{vmatrix}\\&=3(0 \cdot 5-2 \cdot 1)-2(5 \cdot 5-2 \cdot 2)+1(5 \cdot 1-2 \cdot 0)\\&= -6-42+5=-43$

Example C

Find the determinant of $B$  from example B using Sarrus’s Rule.

Solution:

$\begin{matrix}3 & 2 & 1 & 3 & 2\\5 & 0 & 2 & 5 & 0\\2 & 1 & 5 & 2 & 1\end{matrix}$

$\text{det} \ B=0+8+5-0-6-50=-43$

As you can see, Sarrus’s Rule is efficient and much of the calculations can be done mentally.  Additionally, zero values make much of the multiplication easier.

Concept Problem Revisited

Determinants for $2 \times 2$  matrices are defined the way they are because of the general solution to a system of 2 variables and 2 equations.

$ax+by &= e\\cx+dy &= f$

To eliminate the $x$ , scale the first equation by  $c$ and the second equation by a.

$acx+bcy &= ec\\acx+ady &= af$

Subtract the second equation from the first and solve for $y$

$ady-bcy &= af-ec\\y(ad-bc) &= af-ec\\y &= \frac{af-ec}{ad-bc}$

When you solve for $x$  you also get $ad-bc$  in the denominator of the general solution.  This pattern led people to start using this strategy in solving systems of equations.  The determinant is defined in this way so it will always be the denominator of the general solution of either variable.

Vocabulary

The determinant of a matrix is a number calculated from the entries in a matrix.  The procedure is derived from solving linear systems.

Sarrus’s rule is a memorization technique that enables you to compute the determinant of $3 \times 3$  matrices efficiently.

Guided Practice

1. Find the determinant of the following matrix.

$C=\begin{bmatrix}-4 & 12\\1 & -3\end{bmatrix}$

2. Find the determinant of the following matrix.

$D=\begin{bmatrix}4 & 8 & 3\\0 & 1 & 7\\12 & 5 & 13\end{bmatrix}$

3. Find the determinant of the following $4 \times 4$  matrix by carefully choosing the row or column to work with.

$E=\begin{bmatrix}4 & 5 & 0 & 2\\-1 & -3 & 0 & 3\\4 & 8 & 1 & 5\\-3 & 2 & 0 & 9\end{bmatrix}$

1.  $\text{det} \ C=\begin{vmatrix}-4 & 12\\1 & -3\end{vmatrix}=12-12=0$

2.  $\text{det} \ D=\begin{vmatrix}4 & 8 & 3\\0 & 1 & 7\\12 & 5 & 13\end{vmatrix}=4 \cdot 13+8 \cdot 7 \cdot 12+0-36-5 \cdot 7 \cdot 4-0=548$

3. Notice that the third column is made up with zeros and a one.  Choose this column to make up the coefficients because then instead of having to evaluate the determinant of four individual $3 \times 3$  matrices, you only need to do one.

$\begin{vmatrix}4 & 5 & 0 & 2\\-1 & -3 & 0 & 3\\4 & 8 & 1 & 5\\-3 & 2 & 0 & 9\end{vmatrix} &= 0 \cdot \begin{vmatrix}-1 & -3 & 3\\4 & 8 & 5\\-3 & 2 & 9\end{vmatrix}-0 \cdot \begin{vmatrix}4 & 5 & 2\\4 & 8 & 5\\-3 & 2 & 9\end{vmatrix}+1 \cdot \begin{vmatrix}4 & 5 & 2\\-1 & -3 & 3\\-3 & 2 & 9\end{vmatrix}-0 \cdot \begin{vmatrix}4 & 5 & 2\\-1 & -3 & 3\\4 & 8 & 5\end{vmatrix}\\&= \begin{vmatrix}4 & 5 & 2\\-1 & -3 & 3\\-3 & 2 & 9\end{vmatrix}\\&= 4 \cdot (-3) \cdot 9+5 \cdot 3 \cdot (-3)+2 \cdot (-1) \cdot 2-18-24-(-45)\\&= -154$

Practice

Find the determinants of each of the following matrices.

1. $\begin{bmatrix}4 & 5\\2 & 3\end{bmatrix}$

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

3.  $\begin{bmatrix}-1 & 2\\2 & 0\end{bmatrix}$

4.  $\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$

5.  $\begin{bmatrix}6 & 5\\2 & -2\end{bmatrix}$

6.  $\begin{bmatrix}1 & 2\\6 & 3\end{bmatrix}$

7.  $\begin{bmatrix}-1 & 3 & -4\\4 & 2 & 1\\1 & 2 & 5\end{bmatrix}$

8.  $\begin{bmatrix}4 & 5 & 8\\9 & 0 & 1\\0 & 3 & -2\end{bmatrix}$

9.  $\begin{bmatrix}0 & 7 & -1\\2 & -3 & 1\\6 & 8 & 0\end{bmatrix}$

10.  $\begin{bmatrix}4 & 2 & -3\\2 & 4 & 5\\1 & 8 & 0\end{bmatrix}$

11.  $\begin{bmatrix}-2 & -6 & -12\\-1 & -5 & -2\\2 & 3 & 4\end{bmatrix}$

12.  $\begin{bmatrix}-2 & 6 & 3\\2 & 4 & 0\\-8 & 2 & 1\end{bmatrix}$

13.  $\begin{bmatrix}2 & 6 & 4 & 6\\0 & 1 & 0 & 1\\2 & 4 & 2 & 0\\-6 & 2 & 3 & 1\end{bmatrix}$

14. $\begin{bmatrix}5 & 0 & 0 & 1\\2 & 1 & 8 & 3\\9 & 3 & 2 & 6\\-4 & 2 & 5 & 1\end{bmatrix}$

15. Can you find the determinant for any matrix?  Explain.

16. The following matrix has a determinant of zero:  $\begin{bmatrix}6 & 4\\3 & 2\end{bmatrix}$ .  If the determinant of a matrix is zero, what does that say about the rows of the matrix?

Vocabulary Language: English

determinant

determinant

The determinant is a single number descriptor of a square matrix. The determinant is computed from the entries of the matrix, and has many properties and interpretations explored in linear algebra.
Sarrus’ rule

Sarrus’ rule

Sarrus’ rule is a memorization technique that enables you to compute the determinant of matrices efficiently.
Square matrix

Square matrix

A square matrix is a matrix in which the number of rows equals the number of columns.