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Linear Algebra I 1016-
Test 1 Solution
- (6p) Let u =
ļ£» and^ v^ =
ļ£». Find proj^ uv, the projection of^ v^ on^ u.
We compute:
proj (^) uv =
u Ā· v u Ā· u
u =
ļ£» =^
- (6p) Find the complete solution of the following linear system of equations:
x + 2y + 3w = 2 2 x + 4y + 2z + 4w = 6 x + 2y + 2z + w = 4.
We compute a row echelon form for the augmented matrix of the system: ļ£®
ļ£°
ļ£» R^2 ā^2 R ;^1 ,R^3 āR^1
ļ£» R^3 ;āR^2
We use the free variables y and w to find the pivot variables: z = 1 + w and x = 2 ā 2 y ā 3 w. So the complete solution is
ļ£® ļ£Æ ļ£Æ ļ£°
x y z w
ļ£» +^ y
ļ£» +^ w
- (6p) Suppose that u, v ā Rn^ and ||u|| = 1, ||u + 3 v|| = 4, and ||v|| =
- Find the angle between u and v.
First, we compute u Ā· v:
16 = (u + 3v) Ā· (u + 3v) = u Ā· u + 6u Ā· v + 9v Ā· v = 1 + 6u Ā· v + 18,
so u Ā· v = ā 1 /2. This implies that
cos Īø =
u Ā· v ||u||||v||
The angle we are looking for is Īø = arccos(ā 1 /
- (6p) What is the rank of the following matrix?
A =
Use Gauss elimination to create a row echelon form:
A =
R2+R 1 ,R 3 ā 2 R 1 ,R 4 ā 3 R 1 ;
R3+R 2 ,R4+7R 2 / 4 ;
As we can see, the rank is 2.
- (6p) Are the following vectors linearly independent? Explain. ļ£® ļ£Æ ļ£Æ ļ£°
This shows that if a ā 2 b + c 6 = 0, then we have a pivot in the last column, and then the system is inconsistent, so for these vectors in R^3 there is no solution x, y, z, w meaning that these vectors are not in the span of the listed ones (this happens for
example for the vector
ļ£»). The answer is no.
- (6p) Solve the system of equations (a is some real number).
ax + y + z = 1 x + ay + z = 1 x + y + az = 1
For what values of a does the system fail to have solutions, and for what values of a are there infinitely many solutions?
Consider the augmented matrix of the system and create a row echelon form: ļ£®
ļ£°
a 1 1 1 1 a 1 1 1 1 1 a
ļ£» R^1 ;āR^3
1 1 a 1 1 a 1 1 a 1 1 1
ļ£» R^2 āR^1 ;,R^3 āaR^1
1 1 a 1 0 a ā 1 1 ā a 0 0 1 ā a 1 ā a^2 1 ā a
ļ£» R3+ ;R^2
1 1 a 1 0 a ā 1 1 ā a 0 0 0 2 ā a ā a^2 1 ā a
From here we can see immediately that if a = 1 , then this matrix turns into ļ£®
ļ£°
which shows the system will have infinitely many solutions in this case, in the form ļ£®
ļ£°
x y z
ļ£» (^) + y
ļ£» (^) + z
Now if a 6 = 1, then we can divide the second and third row by 1 ā a and we obtain the row echelon form (^) ļ£®
ļ£°
1 1 a 1 0 ā 1 1 0 0 0 2 + a 1
This will show that if a = ā2 , then we have no solutions , and if a 6 = ā 2 (and of
course a 6 = 1), then we have the unique solution
ļ£® ļ£°
x y z
1 /(a + 2) 1 /(a + 2) 1 /(a + 2)
- (6p) Show that for any collection a 1 , a 2 ,... , an of real numbers
a 1 + a 2 +... + an n
a^21 + a^22 +... + a^2 n n
When are the two sides equal?
Use the Cauchy-Schwarz inequality for the vectors
u =
a 1 a 2 .. . an
and v =
1 /n 1 /n .. . 1 /n
The two sides are equal when a 1 = a 2 = a 3 =... = an ā„ 0.
- (6p) Suppose that u, v and w are linearly independent vectors. For what values of k (a real number) are the vectors v ā u, k w ā v and u ā w linearly independent?
Suppose that c 1 (v ā u) + c 2 (kw ā v) + c 3 (u ā w) = 0.
This means that
(c 1 ā c 2 )v + (c 3 ā c 1 )u + (kc 2 ā c 3 )w = 0.
We know that u, v and w are linearly independent, so this means that
c 1 ā c 2 = 0 āc 1 + c 3 = 0 kc 2 ā c 3 = 0.
