# 1. Determinants

by M. Bourne

Before we see how to use a matrix to solve a set of simultaneous equations, we learn about determinants.

A **determinant** is a square array of numbers (written
within a pair of vertical lines) which represents a certain sum
of products.

Below is an example of a 3 × 3 determinant (it has 3 rows and 3 columns).

`|(10,0,-3),(-2,-4,1),(3,0,2)|`

The result of multiplying out, then simplifying the elements of a determinant is a single number (a **scalar** quantity).

## Calculating a 2 × 2 Determinant

In general, we find the value of a 2 × 2 determinant with elements *a*,* b*,* c*,* d *as follows:

`|(a,b),(c,d)|=ad-cb`

We multiply the diagonals (top left × bottom right first), then subtract.

### Example 1

`|(4,1),(2,3)|`

`=4xx3-2xx1`

` =12-2`

` = 10`

The final result is a single **number**.

We will see how to expand a 3 × 3 determinant below.

## Using Determinants to Solve Systems of Equations

We can solve a system of equations using determinants, but it becomes very tedious for large systems. We will only do 2 × 2 and 3 × 3 systems using determinants.

## Cramer's Rule

The solution (*x*, *y*) of the system

`a_1x+b_1y=c_1`

`a_2x+b_2y=c_2`

can be found using determinants:

`x=|(c_1,b_1),(c_2,b_2)|/|(a_1,b_1),(a_2,b_2)|`

`y=|(a_1,c_1),(a_2,c_2)|/|(a_1,b_1),(a_2,b_2)|`

### Example 2

Solve the system using Cramer's Rule:

x− 3y= 62

x+ 3y= 3

Answer

First we determine the values we will need for Cramer's Rule:

*a*_{1} = 1
*b*_{1} = −3
*c*_{1} = 6

*a*_{2} = 2
*b*_{2} = 3
*c*_{2} = 3

`x=|(6,-3),(3,3)|/|(1,-3),(2,3)|=(18+9)/(3+6)=3`

`y=|(1,6),(2,3)|/|(1,-3),(2,3)|=(3-12)/(3+6)=(-9)/9` `=-1`

So the solution is `(3, -1)`.

**Check:**

[1] `3 + 3 = 6` OK

[2] `6 - 3 = 3` OK

**3 ****×**** 3 Determinants**

A 3 × 3 determinant

`|(a_1, b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|`

can be evaluated in various ways.

We will use the method called "expansion by minors". But first, we need a definition.

## Cofactors

The 2 × 2 determinant

`|(b_2,c_2),(b_3,c_3)|`

is
called the **cofactor** of *a*_{1 }for the 3 × 3 determinant:

`|(a_1, b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|`

The cofactor is formed
from the elements that are not in the same row as *a*_{1} and not in the same column as *a*_{1}.

Similarly, the determinant

`|(b_1,c_1),(b_3,c_3)|`

is called the **cofactor** of *a*_{2}. It is formed from the
elements not in the same row as *a*_{2} and not in the same column as *a*_{2}.

We continue the pattern for the cofactor of *a*_{3}.

## Expansion by Minors

We evaluate our 3 × 3 determinant using expansion by minors. This involves multiplying the **elements** in the first column of the determinant by the **cofactors** of those elements. We subtract the middle product and add the final product.

`|(a_1, b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|` `=a_1|(b_2,c_2),(b_3,c_3)|` `-a_2|(b_1,c_1),(b_3,c_3)|` `+a_3|(b_1,c_1),(b_2,c_2)|`

**Note** that we are working down the first column and multiplying by the cofactor of each element.

### Example 3

Evaluate

`|(-2,3,-1),(5,-1,4),(4,-8,2)|`

(You can explore what this example is really asking in this 3D interactive systems of equations applet.)

Answer

` |(-2,3,-1),(5,-1,4),(4,-8,2)|` ` =-2|(-1,4),(-8,2)|` ` -5|(3,-1),(-8,2)|` ` +4|(3,-1),(-1,4)|`

`= -2[(-1)(2) − (-8)(4)] − 5[(3)(2) ` `{: − (-8)(-1)] ` `+ 4[(3)(4) ` `{: − (-1)(-1)]`

`= -2(30) − 5(-2) + 4(11)`

`= -60 + 10 + 44`

`= -6`

Here, we are **expanding by the first column***.* We can do the expansion by using the first row and we will get the same result.

## Cramer's Rule to Solve 3 × 3 Systems of Linear Equations

We can solve the general system of equations,

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}a_{3}x+b_{3}y+c_{3}z=d_{3}

by using the determinants:

`x=|(d_1, b_1,c_1),(d_2,b_2,c_2),(d_3,b_3,c_3)|/Delta`

`y=|(a_1, d_1,c_1),(a_2,d_2,c_2),(a_3,d_3,c_3)|/Delta`

`z=|(a_1, b_1,d_1),(a_2,b_2,d_2),(a_3,b_3,d_3)|/Delta`

where

`Delta=|(a_1, b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|`

### Example 4

Solve, using Cramer's Rule:

2

x+ 3y+z= 2−

x+ 2y+ 3z= −1−3

x− 3y+z= 0

Answer

`x=| (2,3,1),(-1,2,3),(0,-3,1) |/Delta`

`y=| (2,2,1),(-1,-1,3),(-3,0,1) |/Delta`

`z=| (2,3,2),(-1,2,-1),(-3,-3,0) |/Delta`

where

`Delta=| (2,3,1),(-1,2,3),(-3,-3,1) |` `=2(11)+1(6)-3(7)` `=7`

So

`x = (2(11)+1(6)+0)/7=28/7=4`

`y = (2(-1)+1(2)-3(7))/7` `=-21/7` `=-3`

`z = (2(-3)+1(6)-3(-7))/7` `=21/7` `=3`

Checking solutions:

[1] 2(4) + 3(-3) + 3 = 2 OK

[2] −(4) + 2(-3) + 3(3) = -1 OK

[3] −3(4) − 3(−3) + 3 = 0 OK

So the solution is `(4, -3, 3)`.

Easy to understand math videos:

MathTutorDVD.com

## Determinant Exercises

1. Evaluate by expansion of minors:

`|(10,0,-3),(-2,-4,1),(3,0,2)|`

Answer

` |(10,0,-3),(-2,-4,1),(3,0,2)| ` ` =10|(-4,1),(0,2)|` ` -(-2)|(0,-3),(0,2)|` ` +3|(0,-3),(-4,1)|`

`= 10[(−4)(2) − (0)(1)]` ` + 2[(0)(2) − (0)(−3)]` ` + 3[(0)(1) − (-4)(-3)]`

`= 10(-8) + 2(0) + 3(−12)`

`= −80 − 36`

`= −116 `

2. Solve the system by use of determinants:

x+ 3y+z= 42

x− 6y −3z= 104

x− 9y+ 3z= 4

Answer

`x=| (4,3,1),(10,-6,-3),(4,-9,3) |/Delta`

`y=| (1,4,1),(2,10,-3),(4,4,3) |/Delta`

where

`Delta=| (1,3,1),(2,-6,-3),(4,-9,3) |`

`= 1(−45) − 2(18) + 4(−3)`

`= −93`

Note: Once we have *x* and *y*, we can find *z*
without using Cramer's Rule.

So

`x=(4(-45)-10(18)+4(-3))/-93` `=(-372)/-93` `=4`

`y=(1(42)-2(8)+4(-22))/-93` `=(-62)/-93` `=2/3`

Using these two results, we can easily find that *z* =
-2.

Checking the solution:

[1] `(4) + 3(2/3) + -2 = 4`

[2] `2(4) - 6(2/3) - 3(-2) = 10`

[3] `4(4) - 9(2/3) + 3(-2) = 4`

So the solution is `(4, 2/3, -2)`.

Easy to understand math videos:

MathTutorDVD.com

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