1)(20 points) Let A = [v1v2v3] with v1 = (2; 1;1)T, v2 = (0;8; 2)T and v3 = (6;5
ID: 2985159 • Letter: 1
Question
1)(20 points) Let A = [v1v2v3] with v1 = (2; 1;1)T, v2 = (0;8; 2)T and v3 = (6;5;1)T and b = (10;3;3)T
.
a) Are the columns of A linearly independent? Is A invertible?
b) Find the general solution of Ax = b.
c) What is a basis and the dimension of Col(A)? Is b in Col(A)?.
d) What is the rank of A and the dimension of the null space of A?
2) (15 points) Let u = (1;1;1)T, v = (3; 1;2)T and V = spanfvg.
a)Find the projection w = projV u of u onto V.
b) Write u as the sum of w and a vector orthogonal to v.
c) Find the distance from u to V.
3) (20 points) Let A = [x1x2x3] with x1 = (1; 1; 1)T , x2 = (0;3;3)T and x3 = (3;2;4)T
a) Use the Gram-Schmidt process to nd an orthogonal basis for V = Col(A).
b) Find the QR factorization of A. Express R in terms of Q and A, but don't
compute it.
c) Use part b) to determine if A is invertible (no computation is required).
4) (20) Let A = [v1v2v3] with v1 = (7; 3;2)T, v2 = (1;3;2)T and v3 = (2;6;2)T
a) Find the eigenvalues eigenvalue(s) of A.
b) Find bases for the corresponding eigenspaces.
c) Diagonalize A (i.e., write it as A = P DP1). Do not compute P^1 Use it
to nd det(A).
Explanation / Answer
for 3.2, Ax = (1, 0, 0, ... ,0)
then x is the first column of A^-1
Ax = (0, 1, 0, ... , 0)
then x is the 2nd column of A^-1
in general, solve for he kth column by solving Ax = b<k> which has 1 in the kth spot, rest 0
construct A^-1 column by column from these.
it is easy to see that AA^-1 = I, as each column does.
3.1 let x be in R^n. we need to show that there exists y in R^n such that yA = x. this shows that x is a linear combination of the columns of A.
consider y = xA^-1
yA = (xA^-1)A = x(A^-1A) = xI = x
3.3 if the columns of A^-1 were linearly dependent, then its determinant would be 0 and it would not be invertible. But it is invertible, with inverse A.
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