00:02 My name is John Obodo,

00:12 and we are looking at today for the mathematics

00:16 and we want to look at the topic matrices and determinants.

00:21 So, and this area we are going to talk more on determinants.

00:26 And the major thing we want to consider here

00:29 is how to find the determinant of a matrix.

00:32 To find determinants of matrices,

00:33 there are two major approaches at this level,

00:35 and these are the cramers rule and the saros rule.

00:39 So, in dealing with a matrix determinants,

00:42 we consider square matrices two by two

00:45 and the square matrices three by three at this level.

00:49 So, considering a two by two matrix,

00:51 when you are given a matrix A,

00:55 is equal to A1, A2, A3, A4.

01:01 The determinants of matrices A,

01:03 determinants of A is given us A determinants,

01:12 which is sentient as A1, A2, A3, A4.

01:17 So, the determinants is carried out

01:21 by doing a kind of cross multiplication.

01:24 So, when we cross multiply these variables,

01:27 with A1 multiplying A4 and subtracting A2 multiplying A3.

01:32 Now, A1, A4 is the leading diagonal.

01:36 So, you also, you always lead the diagonal,

01:38 they subtract the path that is the follower diagonal.

01:42 So, therefore, we have A1, A2, A3, A4,

01:45 the matrices determinants, we're not being given as A1, A4,

01:48 minus A2, A3, this is for two by two.

01:51 So, for example, if you want to solve the determinants

01:54 of this two by two matrix, we can say A is given us 1, 4, 0, 5.

02:01 Now, the determinants of this will be given as 1 multiplied by 5,

02:07 minus 4 multiplied by 0.

02:10 1 multiplied 5 is 5, 4 multiplied 0 is 0.

02:15 So, 5 minus 0 is 5.

02:17 So, this is for two by two matrices.

02:21 We can consider three by three matrices,

02:24 and dealing with three by three matrices,

02:26 the determinants for three by three matrices,

02:28 we need to look at two major aspects,

02:31 which is the Kramas rule and the Saras rule.

02:34 Let's look at the Kramas rule force.

02:36 Today, with the determinant of three by three matrices,

02:39 using the Kramas rule, we apply what we call

02:42 the co-factor technique of dealing with that.

02:46 And the co-factor technique is applied such a way that

02:50 it will give us the re-values or the values of each position,

02:57 with respect to the magnitude of the determinants,

03:01 either positive or negative.

03:03 So, let's look at the following.

03:05 We have a matrix A, it's given us A 11, A 21, A 31,

03:14 and they are A 12, A 22, A 32.

03:19 And here we have A 13, A 23, and A 33.

03:25 So, this is a square matrix, with three by three,

03:29 because they're three by three columns and three rows.

03:32 So, to solve these three columns and three row matrices,

03:36 let's see what we can do by dealing with the determinants.

03:39 So, we have this as the positive.

03:42 We have negatives, we have positive.

03:45 So, taking this as a co-factor or as a determinant in that case,

03:50 we have the first one as matrix A, the determinants is equal to A 11, A 12, A 21, A 31,

04:03 A 12, A 22, A 32, A 32.

04:09 And they're A 13, A 12, sorry, A 23, and A 33.

04:17 So, we pick the first value or the first position.

04:22 And the rest of that, we say that is equal to plus A 11,

04:29 determinants, A 22, A 32, and the A 23, A 33.

04:41 Close, we have minus, positive, negative and positive.

04:46 So, we have minus A 12, brackets, we have A 21, A 31.

04:56 And the A 23, A 33.

05:03 Then we pick again, plus A 28, 13.

05:09 So, we have plus A 13, plus A 13.

05:14 So, A 13, brackets, we have A 21, A 31, and A 22, A 32.

05:26 Then we do cross-partification on each of them.

05:29 We cross-part fly the leading and the follower, diagonal and do a subtraction.

05:35 So, with that, you can get the answer to that.

05:36 So, we look at an example on that using the cram muscle.

05:40 So, what we have just shown there is cram muscle.

05:43 So, let's solve this problem.

05:45 Matrix A is given us 141, 2, 3, 0, 5, 1, 1.

05:53 Determinant. To find the determinant of this matrix, we say the terminant A is equal to 1,

06:01 OK, brackets, we have 30, 11, close the brackets, OK, we did the terminant sign.

06:09 So, we have minus 2, we have 41, 11, close the bracket, plus 5, determinant, 41, 30.

06:23 So, that we give us 1 in brackets when you do your cross-partification.

06:29 We have 3, minus 0, minus 2 in brackets, 4 minus 1, plus 5 in brackets, 0 minus 3.

06:43 So, we simplify for that we have 1 in bracket 3, minus 2 in brackets 3,

06:51 and the plus 5 in bracket minus 3.

06:55 So, we simplify it, we have 3, minus 6, minus 15.

07:02 So, 3 minus 6, minus 15 will be equal to 3 minus 6 is a minus 3.

07:09 Minus 3 minus 15 is equal to minus 18.

07:14 I would say minus 6 minus 15 is minus 21, minus 21 plus 3 will be equal to minus 18.

07:21 So, the terminant of that matrix that is given, that is matrix A141230511 is minus 18.

07:32 So, with that we can also solve all the related examples as shown on the slide.

07:39 So, the next one here is the Saros rule.

07:41 Today with the Saros rule, the Saros rule is balanced like an expansion of a set of numbers

07:47 of variables along a given line of parts.

07:49 By most cases we need to repeat the values of the leading column,

07:55 or maybe the values of the first column and the values of the second column.

08:01 So, let's see how we can do that.

08:03 We can also do that by repeating the values of the first row and the all and the

08:09 value of the second row downwards.

08:11 So, let's see an example on that and we apply it here.

08:16 So, the given matrix A is equal to A again, A21 or A11, A21, A31.

08:27 Next one is A12, A22, A32.

08:36 The next one is A13, A23 and A33, or you say A33.

08:45 So, we repeat the values of the first column as the first column and the second column.

08:53 Repeat it along the fourth and fifth column respectively then we do the Saros rule.

08:59 First we do that, we have A11, A21, A31, A12, A22, A32, A13, A23, A33.

09:16 Then repeating the first column, we have A11, A21, A31.

09:24 And then this one is A12, A22, A32.

09:30 Then we take the leading diagonals.

09:32 We take the leading diagonals.

09:33 So, we have a diagonal here and this way, take the arrow and another one here, then another

09:39 one here.

09:40 Then you take another one from left to right, sorry from left to right, we have this one,

09:48 we have this two and we have this three.

09:50 So, these ones that are moving from right to left with subtract, the ones that are moving

09:55 from left to right.

09:59 So, we have positive, this one should be positive, positive and this one should be negative,

10:04 negative, negative.

10:05 So, therefore, without we can find the determinant of a battery, this is what we call the

10:11 Saros rule technique, the Saros rule technique.

10:16 So, let's see an example using the Saros rule technique.

10:20 So, to solve a problem using the Saros rule technique, we are giving much speed, it is equal

10:25 to 141 again, 131, 011, for example, I want to find the determinant.

10:33 To find the determinant using the Saros rule, we have 141, 131, 011 and the 141, 131.

10:47 Then, we take this, take this, we take this and we take this, then we also take this one

10:54 in this direction, this direction and this direction.

10:57 We have 1 multiplied by 3 multiplied by 1 times 3 times 1, add it plus 1 times 1 times

11:10 1 times 1 plus 1 times 4 times 0.

11:17 So, after getting that, we now subtract the other aspects, minus, we have this, that means

11:26 we have 1 times 3 times 0, okay, plus, we have another one, 1 times 1 times 1, 1 times

11:38 1, and the next one is a plus, 1 times 4 times 1.

11:46 So, we simplify and let's see what we are going to get, and we simplify this further.

11:53 So, that will give us 1 times 3 times 1 is 3, we have 3 plus 1 plus 0, close the bracket

12:03 minus, we have 0 plus 1 plus 4.

12:10 So, when we add this together, we have 4, add this together, we have 5, so 4 subtract 5

12:19 is minus 1, so the determinant here is minus 1 using the Saros rule, so that is a typical

12:27 example of that.

12:28 So, we can also solve our related problem using the Saros rule technique.

12:33 So, the Saros rule technique is quite different from the method of what we call the Kramas rule,

12:41 okay.

12:42 So, we can also find the values of some certain, what we call the co-factor or the minor of

12:49 a given matrix using the determinant technique, we find the co-factor of a given matrix.

12:56 So, if you are looking for the determinant, we can use the Kramas rule, they will find

13:01 the co-factors.

13:02 So, basically, using the co-factor technique, we also find what we call the inverse of a

13:07 matrix, and they also simplify a given set of equations using the inverse technique.

13:14 So, let's see an example on that to look for the inverse of a matrix, to find the inverse

13:20 of a matrix.

13:21 The inverse of a matrix A inverse of matrix A is given by the adjoint of the matrix, adjoint

13:32 matrix A divided by the determinants of the matrix, adjoint A divided by the determinant,

13:39 and adjoint of matrix A is equal to the transpose of the co-factors.

13:44 The transpose of the co-factors, co-factors of the matrix, the transpose of the co-factors.

13:55 So, adjoint of the co-factors, the transpose of the co-factors, let's look at the co-factors

14:01 of a given matrix.

14:02 Let's see, we have matrix A is equal to A matrix 1, 2, 4, 3, I want to find the inverse,

14:12 so I have to find the inverse, we have to look at the co-factors, what are the co-factors?

14:18 To find the co-factors, A co-factors will be equal to matrix C1, C2, C3, C4.

14:29 So, to find the co-factors, we have C1 is equal to C1 half plus minus, so we have to take

14:38 that as a minus plus. So, C1 here will be equal to 3, and the C2, C2 will be equal to minus

14:49 4, C3 will be equal to a C3, that is minus 2, and the C4 will be equal to C4 matrix will

15:02 be equal to 1, as a C3 is equal to 2. So, with that, you can find the value of the, can

15:14 find the value of the transpose. So, let's see how we can get, we have gotten the co-factors

15:20 of A. So, let's write it out, we have A co-factors, A C is equal to C1 is 3, and the C2 is minus

15:31 4, C3 is minus 2, C4 is 1. So, this is the co-factor of matrix A, and to find the transpose,

15:43 A transpose, the transpose which is AC power T is equal to the transpose in that, we have

15:51 3 minus 2, and the minus 4, 1. So, we have the transpose of the co-factor. So, in this

16:00 case, this is the adjoints of matrix A, adjoints A, adjoints of matrix A. So, the adjoints of

16:08 matrix A is the transpose of the co-factors of the matrix. So, with that, we can now find

16:15 the inverse. So, to find the inverse of the matrix, we have to look for the, we have to

16:21 look for the determinants of the matrix. So, the determinant of the matrix, to find

16:26 the determinants A. The determinants A of the matrix, we're not begotten by looking

16:31 at the matrix, which is given as a 1, 2, 4, 3, 1, 2, 4, 3. So, to get the determinants

16:41 we're not be given as follows. So, we have 1, 2, 4, 3, we'll find the determinants, that

16:47 will give us the determinants. We give us 1, what's left by 3, minus 2, what's left by 4.

16:54 So, 1, what's left by 3, we have 3, minus 6, we have minus 3. So, therefore, the inverse

17:02 of matrix A is 1 of our determinants, multiplied by the adjoints of matrix A. So, we know that

17:11 formula as the formula for the inverse. We now say that is equal to 1 over minus 3 in brackets.

17:19 The adjoint of the matrix is equal to, we have 3 minus 2, minus 4, 1. So, we have a

17:27 3, 3 minus 2. Okay, let's take it, we have 3, minus 2, minus 4, 1. So, this is the inverse

17:39 of matrix 2 by 2. Okay, so that is following the process. So, we can also apply the same process

17:46 in solving 3 by 3 matrix. But, however, you need to also look for the co-factors of the

17:51 matrix as shown. So, we can also apply it in solving a set of linear equations using

17:58 the, what's called the inverse of the matrix or using the Kramas root to solve a set of

18:05 linear equations. Okay, and that's, we give us a very straightforward method of solving

18:12 the inverse of a matrix. So, solution of 3 variables, equation with 3 variables. Then we

18:19 have a x, y, z as the variables. You can form a matrix equation from it and use what's

18:26 called the Kramas root or what's called the Saros root to solve it or it also is the

18:32 inverse root to solve it. So, we have this example, A, is given us, let's say, okay, before we

18:39 do this, let's see the example on the set of equation, sort of linear equation. We're

18:45 giving this a following equation. We have 2x plus 3y minus 4z equal to 1. And we have x minus

18:57 4y plus 5z is equal to 2. And we have 3x plus y minus 2z is equal to 3. To write this

19:11 equation out in matrix form, we have the formula written as, or the set of equation written

19:18 as, with respect to the variables. That is a coefficient. So, what we have is, we have

19:25 this, take the coefficient, we have 2, 1, 3. We have 3 minus 4, 1. And we have a minus 4,

19:35 we have 5, we have minus 2. Then the variables will not be x, y, z, x, y, z, then the solutions,

19:46 or the answer, we have 1, 2, and 3. So, this is written in the form A, x, and the, maybe

19:55 y or B, A, x, B. So, therefore, to find the solution to this, what we do is, we say that

20:02 the values of x, which is x, that is solution of x, will not be A inverse, A inverse B, A inverse

20:12 B. We're A inverse is gotten by looking at the transpose, talk about the adjoint matrix,

20:21 adjoint A, divided by the determinant of A, which you can follow. So, we call this one

20:27 equation 1, I'll call this equation 2. So, following the procedure that we have learnt

20:33 earlier, we can use it to solve the determinant of the matrix, I'm also finding the inverse.

20:40 So, with that you can solve for the solution. We can also use the Krama's rule. Using the

20:44 Krama's rule, we have to find the values of the determinants for x, y, and z, respectively,

20:51 then henceforth we now find the values of the determinants, that is the values of x. So,

20:58 therefore, what I'm saying in essence is that your value of x will be equal to the determinant

21:03 of matrix A x, all over the determinant of matrix A, and y is equal to the determinant of

21:10 matrix A y, over the determinant of matrix A, and z equal to the determinant of matrix

21:18 A z divided by the determinant of matrix A. So, with this you can solve for the values

21:25 of x, y, z by using the Krama's rule. So, this is called the Krama's rule technique or Krama's

21:33 rule method. So, with this you can go through the other examples, that is also given there

21:43 on the slide. So, with that we find the solution of matrices and solution of set of linear

21:50 equations using the determinants method or using what we call the Krama's rule method.

21:58 Thank you for watching. My name is John, I will be back once again. Till next time.