00:01
To solve this problem, we could simply compute the inverse of the matrix obtained in problem 21, but we can also solve it directly.
00:10
So first, we assume the first vector, the first element of order basis c, which was 1 to x as every 1.
00:27
And then we need to find a vector with two components of c1, c2.
00:36
That is when multiplied by the vectors of order basis b gives us w1.
00:46
So we read c1 multiplied by 7 minus 4x plus c2 5x.
00:59
So what we have, here we need to rewrite 7c1 minus 4c1x plus 5c2x.
01:14
If we look, we have 1 x power 1 and x power 2, we need to equate the same powers of x.
01:21
So here i write based on x minus 4c1 plus 5c2 plus 5c2 plus 7 c1 so here this is equal to negative 2 and 7 to 1 so we have a system of equations that's right here and 4 c1 plus 5 c2 is equal to negative 2 and 7 c1 is equal to 1 so c1 is equal to 1 so c1 is and we replace c1 here and we get c2 is equal to negative to 7th.
02:17
So the vector, first vector relative to basis b is going to be 1 -7th and negative 2 -7th...
