Determinant; Definition, How to Find, Benefits and Examples of Problems – In the world of mathematics, determinants are one of the chapters that make you dizzy, requiring high levels of accuracy and patience. So, before we devour how to determine what the determinants are like, let’s try to remember the matrix chapter first.
A. Definition of Determinant
What is that determination? The determinant is the value that can be calculated from the elements of a square matrix. Remember the square matrix? Yup, squares have the same sides, right? Now a square matrix means the number of columns and rows is the same.
In mathematics, a matrix is an arrangement of numbers, symbols or expressions arranged in rows and columns to form a square.
In linear algebra, determinants are values that can be calculated from the elements of a square matrix. The determinant of matrix A is written with the sign det(A), det A, or |A|. The determinant can be thought of as the scaling factor of the transformation described by the matrix.
If in your heart, then you are idly asking
“what if the number of rows and columns is different, can you find the determinant or not?”
The answer is definitely no, yes!
When you learn about matrices, one of the quantities you will learn is determinants. So again, the determinant of the matrix is the value that can be calculated from the elements of the matrix.
Learn about determinants, through the book Introduction to Linear Algebra which is accompanied by a concept map that you can understand more easily.
B. How to find determinants
Determinant This is a scalar quantity or a quantity that only has magnitude/value. The matrix element in question is a square matrix element.
What is a square matrix? A square matrix is a matrix that has the same number of rows and columns, which can be 2×2 or 3×3
1. The determinant of 2×2
Suppose we have a matrix A whose elements are a,b,c,d which is written like this:
Then we will write the determinant of A:
Det A= |A|
then, how to calculate the determinant? hard or easy? hm… have a look first:
The formula for the determinant is :
|A| = ad-bc
First we cross multiply a by d
So
a xd = ad
Then we also multiply b by c
b xc = bc
So again
|A| = ad-bc
To make it easier, you can play around with it, it’s muddy
2. Determinant 3×3
There are two ways to determine the determinant of 3 x 3, namely the Sarrus method and the minor cofactor method. is the sarrus method easier than the sara minor factor or vice versa the sarrus method is more difficult than the cofactor minor method, ok, we’ll brush it off right away!
a. Sarrus way
After discussing how to determine the determinant of 2 x 2 there is also the determinant of 3 x 3 , with more components it seems that finding the determinant is also longer and longer. Let us understand more clearly step by step.
Let’s see, for example, we have the following matrix A:
How to work out the 3 × 3 determinant using the Sarrus method is as follows:
- The determinant is rearranged by adding 2 rows and 3 columns
- then make a diagonal line to the right (red color) with addition operations then diagonally to the left (green line) with subtraction operations.
- we multiply following the line so that we will get the formula for determining the determinant of a 3 × 3 matrix as follows:
|A| =aei + bfg + cdh – ce g – afh – bdi
b. The minor-cofactor way
Apart from the Sarrus method, there is another way to find the determinant of a 3×3 matrix, namely by using the minor cofactor method.
Compared to Sarrus’s method, it seems that the minor cofactor method is longer and more detailed. We have 3 steps to do
- Find M₁₁, M₁₂, M₃
- Create C₁₁, C₁₂ , C₁₃
- Plug it into the 3×3 determinant formula
We will describe one by one, the method is as follows, for example, there is a 3 × 3 A matrix:
What we will do first is to calculate the minor in advance as follows:
1) Looking for minors
We will find the minor of this series. First we will look for the minor column of the 1st row of the 1st way, we delete the 1st column and the 1st row:
After we delete the 1st column and 1st row, we will get the remainder as M₁₁
Second we will delete 1st row 2nd column to get M₁₂
After removing Then we will get the price of M₁₂
Finally. Third, we delete the 1st row and the 3rd column to get M₁₃
After deleting the 1st row and 3rd column we will get:
To make it more stable, let’s try to find M₂₂
If we want M₂₂ then we will delete the 2nd row and 2nd column to get M₂₂
After removing it we will get M₂₂
Once more complete, we find M₃₃ yu, i.e. delete the 3rd row and 3rd column to get M₃₃
Then we will get M₃₃
2) Looking for cofactors
The way to find cofactors is as follows:
Cᵢⱼ= -1ⁱ⁺ʲ |Mᵢⱼ|
suppose we will find C₁₁
i= 1 j= 1
C₁₁= -1¹⁺¹ | M₁₁ |
we already have M₁₁
then
C₁₁= -1¹⁺¹ | ei – fh |
or we can use the following method,
Above the first line we put a sign (+) then (-) after that (+)
In the second line again we put a sign (-) then (+) after that (-)
then in the third line we put a sign (+) then ( -) after that (+)
If you are looking for C₁₁ , in the 1st row the 1st column costs (+)
C₁₁= +[ M₁₁]|
we’ll see :
then the determinant is
C₁₁= + ( e. i – fh)
if looking for C₁₂ , in the 1st row of the 2nd column the price is (+)
note that the sign above b is ( -)
C₁₂= – |M₁₂ |
we’ll see :
then we will get the determinant
C₁₂= – ( d. i – fg)
Finally, if you are looking for C₁₃ , in the 1st row, the 3rd column, the price is (+)
note that the sign above c is (+)
C₁₃= +[ M₁₃] we see :
then the determinant is
C₁₃= + ( dh- eg)
c). looking for determination
Now we come to the point of searching for this very long determinant, namely using a minor cofactor. this is the formula
| A| = A₁₁. C₁₁ + A₁₂ . C₁₂ + A₁₃. C₁₃
we see that
the price of A₁₁ = a
A₁₂ = b
A₁₃ = c
so
| A| = a. C₁₁ + b. C₁₂ + c. C₁₃
Here is the formula for determining the 3 × 3 determinant that has shaken the brain, pounded the heart and made a sudden stomach ache.
You can learn about matrix and linear algebra to better understand determinants through the Matlab book for Linear Algebra and Matrix which explains the concept.
C. Learning Benefit of Determinant Matrix
Maybe you think, what is the use of learning the matrix? Is it just for the sake of getting grades from the math teacher? Eits, don’t get me wrong, the existence of a matrix is enough to help engineers to solve problems that have quite a lot of variables.
In human life the matrix functions or is useful to make it easier to work on data to solve a problem related to numbers and the amount of data collection. the use of matrices usually occurs in table data. For example, for making journals and making reports.
Now matrix theory is usually used to add up the columns in the table as well as subtract, multiply, and divide the values in these columns.
In conclusion, here are some of the uses of matrices in everyday life
- Facilitate in making an analysis of an economic problem that contains a variety of variables
- Used in solving the problem of investigation operations, for example
the problem of investigation of petroleum sources and so on. - Associated with the use of linear programming, input output analysis both in economics, statistics, and in the fields of education, management, chemistry, and other technological fields.
D. Examples of determinants
1. An example of a 2×2 determinant question
The way to understand determinants is, of course, by trying the problem right away, shall we try from the easiest one?
a. Example question 1:
If matrix A is known as follows:
Then determine the determinant
First, because a is represented by 1 d is represented by 3 then ad cross operation 1 with 3 yes
ad = 1×3 = 3
Then because b = 1 c = 2
We multiply
Then bc = 1×2 = 2
So |A| =ad-bc
=(3)-(2)
= 1
so the determinant of A is 1
b. Example question 2
Consider the determinant of matrix B below:
If it is known that the determinant value of matrix B is 4, then the calculated value of x
is ok, we see
|B| = ab- bc
a = 2 b= xc= 4 d = 8
then ad = 2.8 = 16
bc = x.4 = 4x
so |B| = ab- bc
4 = 16 – 4x
4x = 16 – 4
4x = 12
x = 12
4
= 3
so the value of x is 3
c. example question 3:
There are two matrices, namely: matrix A and B as below:
So that the determinant of matrix A is equal to twice the determinant of B, then the value of x that satisfies is
the discussion:
we find det A
a = xb = 2 c = 3 d= 2x
ad= x.2x = 2x²
bc = 2 . 3 = 6
then the determinn A
|A| = ab- bc = 2x² – 6
we find the determinant B
a = 4 b = 3 c = -3 d= x
ad= 4.x = 4x
bc = 3 . -3 = -9
then the determinant
|B| = ab- bc = 4x – (-9) = 4x + 9
because the determinant of matrix A is twice the determinant of B, so
det A= 2 det B
|A| = 2 |B|
2x² – 6 = 2(4x + 9)
2x² – 6 = 8x + 18
2x² – 8x = 18 + 6
2x² – 8x = 24
2x² – 8x – 24 = 0
x² – 4x – 12 = 0
we find the roots
( x -6 ) (x + 2 ) =0
x – 6 = 0
x = 6
x + 2 = 0
x = -2
then the roots are 6 and -2
d. Example question 4
The matrices A and B are known as below:
if the determinant of matrix A is – 5 calculate the determinant of matrix B
we find the determinant of A
|A| = ad- bc
because the determinant A = 5 then
5 = ad- bc
we find the determinant B
|B|| = 3ad- 3bc
= 3( ad – bc)
ad- bc = A= 5
|B|= 3|A|
and |A| = 3. 5 = 15
so the value of the determinant of B is 15
2. Examples of 3×3 determinants
a. Sarrus way
Example problem 1: Determine the determinant of A
Answer :
we do it slowly because it requires a fairly high accuracy.
|A| = aei + bfg + cdh – ceg – afh – bdi
in this determinant we look at the components yes
a = 3 b = 2 c = 1
d = 1 e = 4 f = 2
g = 5 h = 1 i= 0
a.ei = 3. 4. 0 = 0
b.f. g = 2. .2. 5 = 20
c.dh = 1. 1. 1 = 1
c.eg = 1. 4. 5 = 20
a.fh = 3. 2. 1 = 6
b.di = 2. 1. 0 = 0
|A| = aei + bfg + ceh – ceg – afh – bei
= 0 + 20 + 1 – 20 – 6 -0
= -5
Example problem 2: Determine the determinant of the following matrix using the minor-cofactor method! ?
a= 3 b= 1 c= 2
d= 4 e=5 f=1
g = 2 h= 1 i=2
Answer :
First we find M₁₁. by deleting the 1st and 1st rows
after removing row -1 column 1 we will get M₁₁
then we find the price of C₁₁
a= 3 b= 1 c= 2
d= 4 e=5 f=1
g = 2 h= 1 i=2
C₁₁= +[ M₁₁]|
C₁₁= + ( e. i – fh)
= + ( 5.2 – 1.1)
= + ( 10 – 1)
= 9
first we find M₁₂. by deleting the 1st and 2nd lines
then we will get M₁₂
a= 3 b= 1 c= 2
d= 4 e=5 f=1
g = 2 h= 1 i=2
C₁₂= -[ M₁₂ |
= – ( d. i – fg)
= -(4.2 -1.2)
= -( 8-2)
= -6
then find M₁₃ by deleting the 1st and 3rd row
then we will get M₁₃
a= 3 b= 1 c= 2
d= 4 e=5 f=1
g = 2 h= 1 i=2
C₁₃= + [ M₁₃ |
= + ( d. h – eg)
= +(4.1-5.2)
= +( 4-10)
= -6
we plug it into the determinant formula and
we see again that
the price A₁₁ = a= 3
A₁₂ = b = 1
A₁₃ = c = 2
then earlier we got
C₁₁ = 9
C₁₂ = -6
C₁₃ = -6
then
| A| = A₁₁. C₁₁ + A₁₂ . C₁₂ + A₁₃. C₁₃
= (3.9) +(1.6) +( 2.-6)
= 27 -6-12
= 9
So, that’s an in-depth discussion about the 2×2 determinant and 3×3 determinant, hopefully it can help in working on the questions from the teachers. You have to practice this determinant over and over again until you get used to it.