Math Is Fun Forum

okay, the trick with these is to basically get all the trig functions on one side, and all the numbers on the other.

1.

so

so

this you have to solve just by familiarity. The solution in this case happens to be 45 degrees or pi/4.

I don't have time right now to do all of them, but they're really quite easy. All you really need to know is the following identities:

the Pythagorean identities:

and

whenever you see a squared trig funciton, like sin^2 x or cos^2 x and a 1 present, its a dead giveaway, you have to use one of the Pythagorean identities.

hope this helps.

another way to do it, suppose a and c are nonzero. then
multiply the top row by c and the bottom row by a

(ac bc)
(ac da)

subtract the top row from the bottom row

(ac       bc)
(0  bc - da)

now clearly of ad - bc = 0 then bc - da = -(ac - bc) = -0 = 0 which implies the second row is zero and thus having a nonzero row, the matrix is not invertible.

using a similar procedure, we can show that if a or b is zero, and b or c is non zero, then the matrix can be shown to have a zero row. And if all entries are zero, its trivial.

Mathsy's proof is just as good if you are allowed to use the determinant rule, however I suspect you haven't learned that yet or it would have been obvious.

don't mind me, i'm just practicing.

I'm really happy I managed to figure out this proof last night, turned out it was exactly what my book did and I thought it was kind of cool.

suppose

are linearly dependent.
now suppose we one by one remove every vector in

that can be expressed as a linear combination of the others. When we are done we will have a linearly independent subset of

consisting of m elements, call it 

we have the set

is not empty for the vectors are all non zero so the linearly independent subset contains at least one element.

Now take an element

not in the set 

we have that

but also we have that

can be expressed as a linear combination of

so

where

are the scalers of the linear combination. But also by the distributive property of matrix multiplication:

thus

so

since

are linearly independent, we have that all scalers are zero.
Recall now that we have all eigenvalues are unique, and

is not an element of

so

thus 

thus, since they are linearly dependent, and by the zero factor theorem, all scalers

must be zero. But this implies that

and we were given all vectors were nonzero. Hence the contradiction.
So it must be linearly independent.

i can't believe it took me like five minutes of scratching my head to get that one. Hehehe!

agreed. It just sounds weird to say infinity is smaller than a microwave.

1. Woman dead
2. No she was not a scientist.
3. No she was not involved in the arts.
4. No she was not involved with feminism
5. No she was not involved with law.

are r and θ regarded as constants in this problem? or do they have the relationship x = r cos(θ)?

try this. Represent the elements of B as variables a, b, c and d like so
|a b|
|c d|

square this matrix to obtain
|(a^2 + bc)   (ab + db)|
|(ac + dc)   (bc + d^2)|

now you need to find any a,b, c and d such that all these entries become zero. Can you think of any? (hint, use as many zeros as possible, but don't use all zeros, your book probably wants a non zero matrix).

here's how.

show AB is invertible if A and B are invertible.

Well we know A and B have an inverse so if we multiply AB on the right by (B^-1) (A^-1) we have AB(B^-1)(A^-1) = AI(A^-1) = A(A^-1) = I.

now show that AB is not invertible if A and B are not invertible.

we have A and B are not invertible, now suppose AB is invertible, then there is some matrix M such that
(AB)M = I, but this means that
A(BM) = I, which means A has an inverse 'BM' which is a contradiction. Therefore, AB cannot have an inverse, and is not invertible.

welcome to the boards!

So how many programming languages do you know?

1. first use polynomial division to divide 2x^3 - 3x^2 - 11x + 6 by (2x-1), once you're done, if there is no remainder, you have the quietnet is P, that is (2x^3 - 3x^2 - 11x + 6)/(2x-1) = P, multiplying both sides by (2x - 1) you obtain

2x^3 - 3x^2 - 11x + 6 = P(2x-1), your job will then be to factor the polynomial P you got from the division. This will be a quadratic polynomial so it should be easy to factor.

for 2. again divide  4x^3 +12x^2 -7x - 30 by (x+2), if there is no remainder, then it is a factor. It says show that k = -2 though. You'll have to provide more info there because you haven't told us what k represents.