Re: MATH1231/1241/1251 SOS Thread
I'm stuck on a similar one
How do I find this vector???
You essentially have two conditions:
x -(1/4)y + 0z = 0
and
0x -(3/4)y + z = 0
(assuming your answer was right, I didn't check it).
So the vector (x,y,z) is in the column space of A iff it satisfies those two conditions.
You can turn those conditions into a matrix and get it into row-echelon form (actually in this case, it already is in row-echelon form).
Then do the usual procedure of setting a non-leading column's variable to a free parameter (in this case, that variable is z, so set z = lambda say), then use back substitution as usual to get x and y in terms of lambda.
This will mean you'll end up with x, y, z in terms of lambda, which means you can get a vector
b as desired. For the sake of example, if you ended up with x = 2lambda, y = -lambda, z = lambda, we'd have (x,y,z) = (2lambda, -lambda, lambda) = lambda (2,-1,1), and thus a vector we could choose for
b would be (2,-1,1).
\\ \frac{dv(1+v^2)}{v(3+v^2)}&=-\frac{dx}{x}\end{align*})
$ dollars per year. If interest is being paid into the account at a nominal rate of $100R\%$ per year, compounded continuously, find the balance $B(t)$ dollars in the account after $t$ years.)
