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Unformatted text preview: MATH 133  Practice Exam (Solutions) x = 3t + 2
y = −2t + 1 , and the point A(1, −1, 2), Find the
1. Given the following line L1 : z=t
equation of the plane perpendicular to L1 and passing by A.
Solution
n = d1 = (3, −2, 1) and A(1, −1, 2) so the equation of the plane is given by
3(x − 1) − 2(y + 1) + (z − 2) = 0 =⇒ 3x − 2y + z = 7. x = 3t + 2
y = −2t + 1 , and the point A(1, −1, 2).
2. Find the distance between the line L1 : z=t Solution
A general point P on the line has coordinates (3t + 2, −2t + 1, t) for this point
to be the closest to point A we should have vector AP = (3t + 1, −2t − 2, t − 2)
perpendicular to the directional vector of the line (3, −2, 1) thus 3(3t + 1) − 2(−2t −
2) + 1(t − 2) = 0 =⇒ t = −5/14 and the distance between point A and the line
is the norm of the vector AP = AP  = (−1/14)2 + (−9/7)2 + (−33/14)2 3. Find the equation of line AB given that A(1, 2, 3) and B (−2, 0, 2).
Solution
The directional vector of the line is vector AB = (−3, −2, −1) and thus the parametric equations of the line are given by x = −3t + 1
y = −2t + 2 z = −t + 3 4. Find a vector perpendicular to the plane ABC given that A(1, 2, 3), B (−2, 0, 2),
and C (2, 4, −1).
Solution
A vector perpendicular to the plane is the cross product of any of two vectors in
the plane AB = (−3, −2, −1) and AC = (1, 2, −4) for example. n = AB × AC =
(10, −13, −4). x = 3t + 2
y = −2t + 1 ﬁnd the intersection of P1
5. Given (P1 ) : 4x − 3y + z = 2 and L1 : z=t
and L1.
Solution
To ﬁnd the intersection of L1 and P1 we replace the equations of the line into that
of the plane to get
1 3
4(3t + 2) − 3(−2t + 1) + t = 2 =⇒ t = − 19
We now replace in the equations of the line to get the point (29/19, 25/19, −3/19). 6. Find the solution of the following system using GaussJordan elimination
4x1 + x2 − 3x3 + x4 = 1
−x1 + 3x2 − x3 + 2x4 = 7
Solution
4 1 −3 1 1
−1 3 −1 2 7 4R2 + R1 → R1 0 13 −7 9 29
−1 3 −1 2 7 1 −3
1 −2 −7
0 13 −7
9 29 −R2 → R2
R2 ↔ R1
R2 /13 → R2 1 −3
1
−2
−7
0
1 −7/13 9/13 29/13 3R2 + R1 → R1 1 0 −8/13 1/13 −4/13
0 13
−7
9
29
So the solution of the system is given by
x1
x2
x3
x4 = −4/13 + 8/13 t − 1/13 s
= 29/13 + 7/13 t − 9/13 s
=t
=s
k x − 2y = 3
have
2x − ky = 4 7. For what value(s) of k does the system
a) No solution.
b) An inﬁnite number of solutions.
c) A unique solution.
Solution
k −2 3
2 −k 4 k
−2 3
1 −k/2 2 1 −k/2 2
k
−2 3 1
0 R2 /2 → R2
R1 ↔ R2
− kR1 + R2 → R2 −k/2
2
− 4) −2k + 3 1
(k 2
2 a) No solution is when k = 2 or k = −2.
2 b) There is no value of k for which the system has an inﬁnite number of solutions.
c) We get a unique solution when k = 2 and k = −2.
8. Given A = 2
4
−1 −3 a) Write A−1 as a product of elementary matrices.
b) Write A as a product of elementary matrices.
Solution
2
4
−1 −3 −1 −3
2
4
13
24 13
01 − R1 → R1 =⇒ E2 = − 2R1 + R2 =⇒ E3 = 13
0 −2 01
10 R1 ↔ R2 =⇒ E1 = −1 0
01 10
−2 1 R2 /(−2) → R2 =⇒ E4 =
− 3R2 + R1 → R1E5 = −
=⇒ E1 1 = −
=⇒ E2 1 =
−
=⇒ E3 1 = 1
0
0 −1/2 1 −3
01 −
=⇒ E5 1 = a) A−1 = E5 E4 E3E2 E1
−
−
−
−
−
b) A = E1 1 E2 1 E3 1 E4 1 E5 1 9. Find the dimension of the following subspaces of Rn 1
2
4
−1
a) span 0 , 1 , 1 , −1 −1
1
−1
−2 1
1
−1
b) span 1 , 0 , 1 1
2
0 1
0 −1
−2R1 + R2 → R2
2
1
1 −4R1 + R3 → R3
a) 4
1 −1 R1 + R4 → R4
−1 −1 −2 3 −1 0
01 10
21 −
=⇒ E4 1 = 10
01 Solution 01
10 13
01 10
0 −2 0 −1 1
3 1
3
−1 −3 0 −1 1
3 0
0
0
0 =⇒ basis = b) 1
0
0
0
1
0
0
0 1 11
1 02
−1 1 0
111
0 −1 1
021
11 1
0 1 −1
02 1
10 2
0 1 −1
00 3 1
0
, 1 0 −1
3 −R1 + R2 → R2
R1 + R3 → R3 − R2 ↔ R2 −R2 + R1 → R1
−2R2 + R3 → R3 =⇒ dimension = 3.
10. Given that abc
def
ghi Find
−a d 2g
a) −b e 2h
−c f 2i
b) 3a
3b
3c
g + 3a h + 3b i + 3c
2d
2e
2f c) abc
2d 2e 2f
def Solution 4 = −2 −R1 → R1
a) We get the matrix by applying the operations 2R3 → R3
transposing A
so the determiant is given by (−2)(−1)(2) = 4 3R1 → R1
R2 ↔ R3
b) We get the matrix by applying the operations
2R3 → R3
R2 + R1 → R2
so the determinant is given by (−2)(3)(−1)(2) = 12. c) The determinant is 0 since the matrix has 2 proportional rows.
11. Find the determinant of the following matrix 1
0
2 −3 0
2
1 −2
1 0 −1
0
1
1 0
A= 0 −2
1 −1 1 0
0
1 −1 0
Solution
1
0
2 −3
2
1 −2
1
−1
0
1
1
0 −2
1 −1
0
0
1 −1 0
0
0
1
0 1
2
=−
−1
0 0
2 −3
1 −2
1
0
1
1
0
1 −1 = −(−2 − 2 + 3) = 1 1 2 −3
1
= − −1 1
0 1 −1 1
12. Find an orthogonal basis for R3 containing the vector −1 2
Solution 1
1
0 −1 , 0 , 1 span R3 we now use the GramSchmidt
The three vectors
2
0
0
process to get an orthogonal basis. 1
u1 = v1 = −1 2 5/6
u2 = v2 − projv2 u1 = 1/6 .
−1/3 u3 = v3 − projv3 u1 − projv3 u2 5 13. Find W⊥ given each of the following cases 1
2 −1 , 0 a) W = span
2
1 1 0 b) W = span 2 −1
Solution 12
a) W = col(A) where A = −1 0 21 1 −1 2
is
2
01 −1/2 =⇒ W⊥ = span −3/2 . 1 1 = col(A) where A = 0 2
−1 W⊥ = null(AT ) The RREF form of
x = −1/2 t
=⇒ y = −3/2 t
z=t 1 0
b) W = span 2
−1
W⊥ = null(AT ) [1 0 2 − 1] =⇒ x = t − 2s
y=r
z=s
w=t −2 0
s + 1
0 x1
1 x2 0 = t + x3 0 1
x4 1 0 −1
14. Given A = 0 1 0 −1 0 1 10
1/2
0 1 −3/2 0
1
r
0
0 a) Find an orthogonal matrix P and a diagonal matrix D such that D = P T AP .
Solution 1−λ
0
−1
(A − λI ) = 0
1−λ
0
−1
0
1−λ
6 det(A − λI ) = λ(λ − 1)(λ − 2) =⇒ λ = 0, λ = 1, λ = 2 √
1
1/ 2
λ = 0 =⇒ e1 = 0 =⇒ q1 = √0 1
1/ 2 0
0 1 =⇒ q1 = 1 λ = 1 =⇒ e2 =
0
0
√ 1
1/ 2 λ = 2 =⇒ e3 = 0 =⇒ q1 = √0
−1
−1/ 2
√
√
1/ 2 0 1/ 2
0
0√ Q=
√1
1/ 2 0 −1/ 2 000
D= 0 1 0 002 15. Find the matrix of the linear transformation for a) the counterclockwise rotation about the origin by and angle of π .
3
b) the relfection about the line 3x1 − 2x2 = 0.
c) the projection on the line 4x1 − x2 = 0.
Solution
a) T = cos(π/3) − sin(π/3)
sin(π/3) cos(π/3) = √
1/2 − 3/2
√
3/2
1/2 b) The image of (x, y ) under this transformation is given by
7x 24y 24x 57y
,
,
)
(x, y ) − 2proj(3,−4)(x, y ) = ( −
25
25 25 25
7/25 24/25
so T is given by
−24/25 57/25
c) The image of (x, y ) under this transformation is given by
x
4y 4x 16y
proj(1,4)(x, y ) = ( + ,
+
)
17 17 17
17
so T is given by 1
17
4
17 4
17
16
17 16. Are the vectors w,u, and v independant? 1
5
3 −2 , u = 6 , v = 2 .
a) w =
3
−1
1
7 1
2
0 0 , u = −1 , v = 2 .
b) w =
3
1
1
Solution
To solve this problem we write the vectors as the rows (or columns) of a matrix and
then either reduce the matrix to RREF or compute the determinant of the matrix
• if the RREF of the matrix is the identity or the determinand is not zero then
they are indepedant.
• if the RREF of the matrix has at least one row of zeroes or the determinant is
zero then the vectors are linearly dependant. 1 −2 3
a) det 5 6 −1 = 0 =⇒ dependant.
32
1 103
b) det 2 −1 1 = 9 = 0 =⇒ independant.
021 8 ...
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 Spring '08
 KLEMES
 Math

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