Linear Algebra
Test #1
March 21, 2003
Name____________________
R.  Hammack
Score ______


(1) (9 points) This problem concerns the vectors [Graphics:Images/T1S03Asol_gr_1.gif],  [Graphics:Images/T1S03Asol_gr_2.gif],  [Graphics:Images/T1S03Asol_gr_3.gif].
(a) [Graphics:Images/T1S03Asol_gr_4.gif][Graphics:Images/T1S03Asol_gr_5.gif]

(b) [Graphics:Images/T1S03Asol_gr_6.gif](1)(2) + (3)(1) + (2)(3) = 11

(c) Two of the vectors u, v, and w are orthogonal. Which two, and why?
v and w because their dot product is zero.

(2) (10 points) Find the distance between the point [Graphics:Images/T1S03Asol_gr_7.gif]in  [Graphics:Images/T1S03Asol_gr_8.gif] and the plane [Graphics:Images/T1S03Asol_gr_9.gif].
Call the above point q =[Graphics:Images/T1S03Asol_gr_10.gif].  A point on the plane is p = [Graphics:Images/T1S03Asol_gr_11.gif].  A normal vector to the plane is n =  [Graphics:Images/T1S03Asol_gr_12.gif].  The vector pointing from p to q is r = q - p = [Graphics:Images/T1S03Asol_gr_13.gif].  The distance from q to the plane is [Graphics:Images/T1S03Asol_gr_14.gif] [Graphics:Images/T1S03Asol_gr_15.gif].


(3) (10 points) Find the solution of the following linear system. Write the solutions in vector form.

[Graphics:Images/T1S03Asol_gr_16.gif]
[Graphics:Images/T1S03Asol_gr_17.gif]

[Graphics:Images/T1S03Asol_gr_18.gif]

[Graphics:Images/T1S03Asol_gr_19.gif]
[Graphics:Images/T1S03Asol_gr_20.gif]

[Graphics:Images/T1S03Asol_gr_21.gif]
[Graphics:Images/T1S03Asol_gr_22.gif]free
[Graphics:Images/T1S03Asol_gr_23.gif]
[Graphics:Images/T1S03Asol_gr_24.gif]free

Solutions are
[Graphics:Images/T1S03Asol_gr_25.gif]

(4)
(a) (10 points) Consider the set of vectors  [Graphics:Images/T1S03Asol_gr_26.gif]  in [Graphics:Images/T1S03Asol_gr_27.gif].  Is this a line or a plane?  If it is a line, write its equation in vector from. If it is a plane, write its equation in normal form.

Notice the second vector is a multiple of the first so it can be eliminated. Then  [Graphics:Images/T1S03Asol_gr_28.gif] = [Graphics:Images/T1S03Asol_gr_29.gif] is line through the origin. It's vector form is [Graphics:Images/T1S03Asol_gr_30.gif].


(b) (10 points) Consider the set of vectors  [Graphics:Images/T1S03Asol_gr_31.gif]  in [Graphics:Images/T1S03Asol_gr_32.gif].  Is this a line or a plane?  If it is a line, write its equation in vector from. If it is a plane, write its equation in normal form.

This time the two vectors are not multiples of each other, so we are dealing with a plane throught the origin.
Its normal form must be ax + by + cz = 0, so we just need to find a b and c. Since the two vectors are on the plane we get the following system.

[Graphics:Images/T1S03Asol_gr_33.gif]

Solving:
[Graphics:Images/T1S03Asol_gr_34.gif]
The solutions are thus a = -c and b = c, with c free. Set c = 1 to get b = 1 and a = -1.

The general form of the plane is thus -x + y + z = 0.
Its normal form is thus [Graphics:Images/T1S03Asol_gr_35.gif]


(5) (10 points) Suppose A, B and X are invertible n-by-n matrices. Solve the following equation for X.

[Graphics:Images/T1S03Asol_gr_36.gif]
[Graphics:Images/T1S03Asol_gr_37.gif]
[Graphics:Images/T1S03Asol_gr_38.gif]
[Graphics:Images/T1S03Asol_gr_39.gif]
[Graphics:Images/T1S03Asol_gr_40.gif]
[Graphics:Images/T1S03Asol_gr_41.gif]
[Graphics:Images/T1S03Asol_gr_42.gif]
[Graphics:Images/T1S03Asol_gr_43.gif]
[Graphics:Images/T1S03Asol_gr_44.gif]
[Graphics:Images/T1S03Asol_gr_45.gif]
[Graphics:Images/T1S03Asol_gr_46.gif]

(6) This problem concerns the vectors [Graphics:Images/T1S03Asol_gr_47.gif], [Graphics:Images/T1S03Asol_gr_48.gif], [Graphics:Images/T1S03Asol_gr_49.gif].

(a) (10 points) Are these vectors linearly independent or linearly dependent? Show your work.
Setting this up in the ususal way, we look at the solutions of the system

[Graphics:Images/T1S03Asol_gr_50.gif]

Setting up the augmented matrix,
[Graphics:Images/T1S03Asol_gr_51.gif]
So you see there's only the trivial solution, so the vectors are linearly independent.


(b) (5 points) Is the span of these vectors equal to [Graphics:Images/T1S03Asol_gr_52.gif]?  Why or why not?

NO. Notice the third component of each vector is 0. Any linear combination of them will thus have a zero in the third component. Since o [Graphics:Images/T1S03Asol_gr_53.gif]has vectors whose third components are not zero, the three vectors in this problem cannot span it.


(7) (10 points) Suppose A is a 3-by-4 matrix. Are the columns of A linearly independent, linearly dependent, or is there not enough information to say?  Explain.

Let A = [a b c d] where a, b, c and d are vectors in [Graphics:Images/T1S03Asol_gr_54.gif]. In testing for independence, we look at the solutions of the system xa + yb + zc +wd = 0. This is a homogeneous system of three equations in 4 variables. Solving it will thus result in free variables, so there will be a nontrivial solution. Thus the columns are linearly dependent.


(8) (10 points) This problem concerns the invertible matrix [Graphics:Images/T1S03Asol_gr_55.gif]
(a) Find [Graphics:Images/T1S03Asol_gr_56.gif].
[Graphics:Images/T1S03Asol_gr_57.gif]

Thus [Graphics:Images/T1S03Asol_gr_58.gif]


(b) (6 points)  Use your answer to part (a) above to find a solution to the equation [Graphics:Images/T1S03Asol_gr_59.gif].

[Graphics:Images/T1S03Asol_gr_60.gif].