00:01
Problem number 2, which an a, we first say that t1 is equal to 1 plus 0x plus 0x squared, so it's equal to 1 minus 3 times 0, and 2 times 1 plus 0 ,000 times 0, which will be equal to 1 and 2.
00:23
For t x, we have here the x coefficient, which is corresponding to 0 minus 3 times 0, and 2 times 0 plus 1 minus 2 times 0, which will be equal to 0 and 1.
00:39
For t x squared, so here we have only x squared coefficient, and this will give negative 3 and negative 2.
00:48
To show that t1 and tx and tx squared are linear combinations.
00:58
So we can think that t1 is equal to 1 times 1 and 0 plus 2 times 0 and 1, and tx is equal to 0 times 1 and 0 plus 1 times 0 and 1, and tx squared is equal to negative 3, 1 and 0 plus negative 2, 0, 0 ,000, and 2, 0, 0 and 1, which will give negative 3 and negative 2 as shown here.
01:21
So we can say that t1 of c is equal to 1 and 2, t of x is equal to 0 1, 2 of x squared is equal to negative 3 and negative 2.
01:33
So that t of c and b is 1 and 2, 0 1, and negative 3, and negative 2.
01:38
Similarly for couchin b, we say that t1, as we saw here, is 1 and 2, the same as here.
01:47
For t1 plus x, here we have post -composition 1 and x...