00:01
We now write the system of linear equations as a matrix equation.
00:05
The matrix equation that we write is ax equals b, where a represents the coefficient matrix, and x represents the matrix of unknowns, and b represents the constant matrix.
00:19
So first, i'm going to create the a matrix by using the coefficients in each of the equation.
00:26
So when we construct the first equation, the coefficients are 1.
00:31
Negative 2 and 3 so we put this as the first row of the a matrix and similarly if we consider the second equation the coefficients are negative 1 3 and negative 1 and for the third equation the coefficients are 2 negative 5 and 5 so this forms the a matrix which is the coefficient matrix we like this in matrix form and then we replace this x matrix by the matrix of unknowns.
01:04
Here the unknowns are x1, x2 and x3.
01:08
So we basically put this in a column format that is x1, x2 and x3.
01:14
This represents the unknown matrix that is x and this equals we put the constant matrix which is basically the constants that are available on the right side of each of the equations.
01:28
So the constant matrix is next.
01:30
Negative 6, 6 and negative 13, we put this as a column matrix b.
01:37
And so we have represented the given system of linear equation in matrix performance.
01:48
Now let's answer the part b question.
01:50
Here we have to solve for matrix x using the goss jordan elimination.
01:55
So the first thing that we have to do in goss jordan elimination method is we have to construct an augmented matrix for the system that is, given to us so we have already represented the system of equations in matrix form we will utilize that to construct the augmented matrix first we write this a matrix as it is that is 1 negative 2 3 in the first row then negative 1 3 negative 1 in the second row and then 2 negative 5 5 in the third row and we also put a line next to this and now include the constant matrix that is we have to include only this column matrix as the cluster column of this augmented matrix and that is negative 6 6 and negative 13 so this basically forms the augmented matrix which is required in gosh certain elimination method to solve for the matrix x the next step is to perform sequence of row operations on this augmented matrix so that at one stage we will end up with a form called row reduced echelon form, shortly written as rref.
03:35
This basically stands for a row reduced echelon form.
03:49
So once we convert the augmented matrix into row reduced echelon form, we can easily determine the matrix of unknowns and thereby solve the solution to the system of equation.
04:05
And so let's start off with the augmented matrix over here.
04:10
The first thing that we have to do is in a, if you do the row operations, we basically have to keep the diagonal elements as it is or we can make this diagonal elements as non -zero.
04:24
So here the diagonal elements all are non -zero.
04:27
And i'm going to make this elements as 0.
04:32
We have to first make this elements as 0.
04:36
And so for that, we perform the below operations.
04:41
That is, we write r2 equals r2 plus r1.
04:52
That is we add these two rows.
04:55
When we add these two rows, this negative 1, this positive 1 will become 0.
04:59
And that is why we perform this row operation.
05:03
And similarly, to make this element 0, that is, we basically convert r3.
05:10
So that is r3 is r1 plus we multiply this r1 by negative 2.
05:21
If we multiply this by negative 2 and add with 2, we get 0.
05:25
So we basically perform negative 2 times of the row 1 and then add with row 3.
05:31
So this will be new r3.
05:34
And this means if we can write the first row as it is, the first row elements are 1, negative 2, 3.
05:42
We also have negative 6.
05:47
And then we add the first row with the second row.
05:53
So i'm going to write the first row elements here, which is 1, negative 2, 3 and negative 6.
05:59
Write the second row elements which is negative 1, 3, negative 1 and 6.
06:06
We add these 2.
06:07
This will be 0.
06:10
This is 1 and this is 2 and this is 0.
06:13
So here we get 0 1 to 0.
06:17
So this will be new or 2 elements.
06:20
So i'm going to write this here.
06:22
0 1, 2 .0.
06:24
And then we perform this operation that is negative 2 times of our.
06:29
1 plus r 3 so first let me multiply this row 1 by negative 2 each of the elements has to be multiplied by negative 2 which means we can write that as negative 2 negative 2 times negative 2 is positive 4 this will become negative 6 this will become 12 and then we add with row 3 elements the row 3 elements are 2 negative 5 5 5 and negative 13 is perform the solution so these two will become zero here it is negative one and here also negative one and here also negative one so we basically get zero negative one negative one negative one this is our new r3 so here i write the new or three elements which is zero negative one negative one and negative one we also put this line to separate the coefficient matrix with a constant matrix.
07:37
Now we can see that we have converted these elements as zero.
07:41
This also converted as zero.
07:43
Now the next thing is we have to convert this as zero.
07:47
And for that we perform this row operation.
07:50
That is we add r3 with r2.
07:57
If we add these two, they will become 0 so this will become the new r3 this to that i'm going to write the first two elements first two row elements as it is that is 1 negative 3 i'm sorry negative 2 3 negative 6 and also write the second row element as it is 0 1 2 0 because only r 3 is getting transponed so now we just have to add these 2 0 plus 0 is 0, negative 1 plus 1 is 0, and then 2 negative 1, that is positive 1.
08:37
0 plus negative 1 is negative 1.
08:40
So we get this after we performed this row operation.
08:46
So now we can see that we have converted this element to 0.
08:50
Now we should also convert this element to 0 as well as this element 0.
08:57
And for that, we perform this row elements, row operation.
09:02
That is, we transform r2 by multiplying negative 2 with r3 and then adding with r2.
09:13
That is, we take the row 3 elements multiply by negative 2 and add with row 2 elements.
09:21
That will be the new r2.
09:23
Similarly, we also transform r1 by performing this row operation.
09:28
Here we multiply negative 3 r3 and then add with r1.
09:34
So let's perform this first.
09:37
Here i have to multiply the row 3 with negative 2.
09:40
So here we have 0 .0.
09:43
I had to multiply this with negative 2, negative 2 and this will be positive 2...