00:01
We are given with the basis.
00:03
Basis of b are 6, 1, 3 and 1.
00:14
And basis of c are also given.
00:16
That are 3, 1 and negative 3, 0.
00:25
We have to find a matrix that is by the bound b to c.
00:32
We have to change this basis of b into c.
00:35
We will find that matrix that we change b into c.
00:40
Just write the vectors or the basis of b.
00:44
It is 6 and 1.
00:48
We have to write in the form of c basis.
00:51
That mean the c basis are 3, 1 and negative 3, 0.
00:57
There should be also multiple of some number.
01:00
So they are linear combination of them.
01:02
From this we get the two equations.
01:04
6 is equal to 3c1 plus negative 3c2.
01:11
And from second we get 1 is equal to c1.
01:16
And this is 0 into c2 that is 0.
01:18
From here we get c1 is 1.
01:20
If c1 we have find that put c1 is here.
01:24
We get c2 is equal to negative 1.
01:28
For this we find the values of c.
01:30
Now for the other vector of b that is 3 and 1.
01:34
For this we have to write in the form of this.
01:38
We will get c1.
01:40
C1 is a scalar multiple.
01:43
That is unknown value we are finding there.
01:46
3, 1 plus c2 negative 3 into 0.
01:53
Just finding its equation.
01:55
These are 3 is equal to 3c1 negative 3c2.
02:01
From the last equation we get 1 is equal to c1.
02:04
C1 is 1 puts values of c1 is 1.
02:08
We get c2 is equal to 0.
02:11
C1 is 1 and c2 is 0.
02:13
For this matrix p from b to c will be.
02:19
Write the coefficient that we have found.
02:21
In first case 1 and negative 1.
02:24
So write it 1 and negative 1...