MATH 2090 QUIZZES FALL 2000

QUIZ 1

Solve the initial value problem   xy' - 3y = 4x⁴ - x²,  y(1)=0

  y = 4x⁴ + x² - 5x³

QUIZ 2

QUIZ 3

Recall that R³ denotes the vector space of all ordered 3-tuples of real numbers. Let S be the set of all vectors (x₁,x₂,x₃) in R³ satisfing the condition 3x₁-x₂+7x₃=0. Show that S is a subspace of R³.

Check that S is closed under vector addition and under scalar multiplication:

Let u = (a,b,c) and v = (e,f,g) be vectors in S. Then 3a-b+7c=0 and 3e-f+7g=0. We must show that u+v=(a+e,b+f,c+g) is in S. This is the case if 3(a+e)-(b+f)+7(c+g)=0.

3(a+e)-(b+f)+7(c+g)=3a+3e-b-f+7c+7g=(3a-b+7c)+(3e-f+7g)=0+0=0

So u+v is in S, and S is closed under vector addition.

Let u=(a,b,c) be in S as above, so 3a-b+7c=0, and let k be a scalar. We must show that ku=(ka,kb,kc) is in S. This is the case if 3(ka)-(kb)+7(kc)=0.

3(ka)-(kb)+7(kc)=k(3a-b+7c)=k(0)=0

So ku is in S, and S is closed under scalar multiplication.

Since S is closed under vector addition and under scalar multiplication, S is a subspace of R³.

The subspace S is the plane through the origin in R³ consisting of all vectors x=(x₁,x₂,x₃) for which 3x₁-x₂+7x₃=0. So x₂=3x₁+7x₃, and

x=(x₁,3x₁+7x₃,x₃)= x₁(1,3,0)+x₃(0,7,1)

Thus every vector in S can be written as a linear combination of the vectors (1,3,0) and (0,7,1). So {(1,3,0),(0,7,1)} is a spanning set for S.

QUIZ 4

				|	2	-3	2	|
The matrix		A	=	|	-1	0	2	|		has characteristic polynomial		p(r) = (1-r)(r-2)(r-2).
				|	-1	-3	5	|

Find a nonsingular matrix S and a diagonal matrix D so that S^-1AS = D.

One answer is

		|	1	-3	2	|				|	1	0	0	|
S	=	|	1	1	0	|		D	=	|	0	3	0	|
		|	1	0	1	|				|	0	0	3	|

QUIZ 5

Find a fundamental set of solutions for the system	x' = Ax	for the matrix	A	=	|	2	-3	|
					|	-1	4	|

One fundamental set of solutions is { x₁ = e^5t(-1,1), x₂ = e^t(3,1) }
(eigenvectors written in "in-line" notation).