00:01
So to solve this problem, the first thing i'd like to note is that these transformations are linear, which means i can distribute the t over to 2v1 and distribute it over to 3v2.
00:14
In the similar, i can distribute t over v1 and distribute it over v2.
00:19
So our expressions become t of 2v1 plus t of 3v2, which is equal to v1 plus v2, and in similar vein, it's equal to t of v1 plus t of v2, which is equal to 3v1 minus v2.
00:42
And then secondly, because t is linear, i can pull out my constant term of 2 and the constant term of 3.
00:50
So this expression becomes 2 times t of v1 plus 3 times t of v2, which is then equal to v1 plus v2.
01:03
So in order to solve for the system of equations, i'm going to subtract from this equation from this one by multiplying it by two.
01:15
So it goes like this...