00:01
The answer to this problem is the inverse of the matrix obtained in problem 17, but to solve it separately, we first write the first vector of basis c equals to, and then we need to find a vector c1, c2, that then multiply by the vectors of b gives us the first first first.
00:33
Vector of c.
00:35
And i mean here we write c9 and 2 plus c2 for negative 3.
00:50
So here this is x, this is y.
00:58
And then c1 multiplied by the x of the first and c2 multiply the x of the second gives us x.
01:08
So system of equations here, 9c1 plus 4c2 gives us 2, and 2c1 plus here is negative.
01:23
So minus 3c2 gives us 1.
01:31
Then we need to solve this by just simplifying it.
01:36
Here we can write c1 is equal to 1 plus 3, c2 divided by 2, and then replace c1 here, and then we can get c1 and c2, which is going to be write it as w, relative to the basis b to 7th, and negative negative 1, 0...