UBC Mathematics 152 Section 206
Midterm Exam II (Mar 16, 2012)
Instructor: Yue-Xian Li
Duration: 50 min
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I.
(30 marks)
(1) Determine if each of the following is a linear transformation. Justify briefly.
(a) f (x, y, z) = (x − y, 3y, 3x + 2z).
(b) g(x, y, z) = (x + y, 3xy, y − z).
(2) Find the corresponding matrix M of each of the linear transformation found
in (1).
(3) For each linear transformation, determine if it is invertible. If yes, find M −1 ;
if not, briefly explain why it is not.
(4) For each linear transformation, find x if M x = b, where x = [x y z]T and
b = [1 0 − 1]T .
Solutions:
(1)
(a) f (x, y, z) = (x − y, 3y, 3x + 2z) is a linear transformation because
f (sx, sy, sz) = (sx − sy, 3sy, 3sx + 2sz) = s(x − y, 3y, 3x + 2z) = sf (x, y, z);
and
f (x1 +x2 , y1 +y2 , z1 +z2 ) = (x1 +x2 −(y1 +y2 ), 3(y1 +y2 ), 3(x1 +x2 )+2(z1 +
z2 )) = (x1 − y1 , 3y1 , 3x1 + 2z1 ) + (x2 − y2 , 3y2 , 3x2 + 2z2 ) = f (x1 , y1 , z1 ) +
f (x2 , y2 , z2 ).
Or equivalently, f (sx1 + tx2 , sy1 + ty2 , sz1 + tz2 ) = (sx1 + tx2 − (sy1 +
ty2 ), 3(sy1 + ty2 ), 3(sx1 + tx2 ) + 2(sz1 + tz2 )) = s(x1 − y1 , 3y1 , 3x1 + 2z1 ) +
t(x2 − y2 , 3y2 , 3x2 + 2z2 ) = sf (x1 , y1 , z1 ) + tf (x2 , y2 , z2 ).
(b) g(x, y, z) = (x+y, 3xy, y−z) is not a linear transformation because g(sx, sy, sz) =
(s(x + y), 3s2 xy, s(y − z)) = s(x + y, 3sxy, y − z) 6= sg(x, y, z).
1
 
 
1
−1
(2) f (e1 ) = f (1, 0, 0) = 0 ,
f (e2 ) = f (0, 1, 0) =  3  ,
f (e3 ) =
3
0
 


0
1 −1 0
f (0, 0, 1) = 0
⇒
M = [f (e1 ) f (e2 ) f (e3 )] = 0 3 0 .
2
3 0 2
1 −1
(3) For the matrix M found in (2), det M = 2 det
= 2 × 3 = 6 6= 0. So, M
0 3
must be invertible.
M −1 can be found by using row operations
prefers.

1 −1 0

Using row operations, we obtain 0 3 0
3 0 2



1 0 0 | 1 1/3 0
1 0 0
(3)/2
0 1 0 | 0 1/3 0 −

−−→ 0 1 0
0 0 1
0 0 2 | −3 −1 1




6
2 0
1
1/3
0
1
 0
2 0.
1/3
0 =  0
6
−9 −3 3
−3/2 −1/2 1/2
or the co-factor formula whichever one
|
|
|
|
|
|



1 0 0
1 −1 0 | 1
0 0
(2)/3
(1)+(2)
0 1 0 −−−−−→ 0 1 0 | 0 1/3 0 −−−−−→
(3)−3(1)
(3)−3(2)
0 0 1
0 3 2 | −3 0 1

1
1/3
0
0
1/3
0 
⇒
M −1 =
−3/2 −1/2 1/2

Using the co-factor formula, we obtain M −1

det M11 − det M21 det M31
1 
− det M12 det M22 − det M32  =
=
det M
det M13 − det M23 det M35


6
2 0
1
0
2 0.
6 −9 −3 3

 

  
6
2 0
1
6
1
1
1
2 0  0  =  0  =  0 .
(4) Since M is invertible, x = M −1 b =  0
6 −9 −3 3
6 −12
−1
−2
II.
(30 marks)
Consider a linear transformation T: R2 → R2 that rotates a vector counterclockwise
by 45◦ followed by a reflection in the line y = x.
(a) Find its matrix representation, i.e. find the matrix A such that T (x) = Ax
for each x ∈ R2 .
(b) Find the inverse A−1 and explain this matrix in terms of a composition of a
reflection and a rotation.
(c) If the transformation defined in (a) is further followed by a projection onto
the x-axis, would the resulting transformation be invertible? Justify your
answer.
Solution:
(1) Since T=Ref(45◦ ) ◦Rot(45◦ ) . A simple graphic plot allows us to figure out
that
√ √ 1
1
√2 ,
√2 .
T (e1 ) =
T (e2 ) =
2
2
2 − 2
√ 2 1 1
.
Therefore, T (x) = Ax with A = [T (e1 ) T (e2 )] =
2 1 −1
Alternatively, we can use the
formulas
cos(90◦ ) sin(90◦ )
A = MRef(45◦ ) MRot(45◦ ) =
sin(90◦ ) − cos(90◦ )
√ #
√ "√
2 1 1
cos((45◦ ) − sin((45◦ )
0 1 √22 −√ 22
=
.
=
2
2
sin((45◦ ) cos((45◦ )
1 0
2 1 −1
2
2
(2) Since T=Ref(45◦ ) ◦Rot(45◦ ) , the inverse should be T −1 =(Ref(45◦ ) ◦Rot(45◦ ) )−1
−1
◦
◦
=Rot−1
(45◦ ) ◦Ref(45◦ ) =Rot(−45 ) ◦Ref(45 ) . Therefore, the inverse is a composition of a reflection in y = x followed by a rotation of 45◦ clockwise.
" √
√ #
√ 2
2
2 1 1
0 1
−1
2
2
√
√
.
A = MRot(−45◦ ) MRef(45◦ ) =
=
2 1 −1
− 22 22 1 0
Alternative, we can use the formula
√ √ √ 2 −1 −1
2 −1 −1
2 1 1
1
−1
=−
=
.
A =
det A 2 −1 1
2 −1 1
2 1 −1
(3) No, it won’t be invertible since projection is not one-to-one. Any vector
starting from one fixed point (not the origin) on the x-axis and ending on
different points on the y-axis are projected to the same vector on the x axis.
Alternatively,
cos2 0
cos 0 sin 0
1 0
=
,⇒
MP rojθ=0 =
0 0
cos 0 sin 0
sin2 0
√
√ 2 1 1
2 1 1
1 0
A = MP rojθ=0 MRef(45◦ ) MRot(45◦ ) =
=
,
0 0 2 1 −1
2 0 0
which is not invertible since det A = 0.
III. These questions do not require
the boxes.

0.3
(a) True or false. Matrix M = 0.2
0.5
walk problem.
lengthy calculations. Put your answers in
(16 marks, 4 for each)

0.2 0.4
0.6 0.4 is a transition matrix for a random
0.3 0.2
False. The sum of the entries in the 2nd column is not 1.
(b) True or false. For a rotation matrix A =
1
2
√
3 √
−1
, A−1 = AT .
3
1
True.
(c) Let A = MRot30◦ be the matrix that rotates a 2D vector counterclockwise by
30◦ and B = MRef−45◦ be the matrix that reflects in the line y = −x. Find all
statements that is true in the following list of statements: (i) A3 = I; (ii) A12 = I;
(iii) B 3 = I; (iv) B 12 = I; (v) A−1 = AT ; (vi) B −1 = B.
(ii),(iv),(v),(vi).
(d) For the circuit given in the figure, assume that V and I are known. Find the
current that flows through the resistor that is equal to 2Ω and E (express them in
terms of V and I).
Left loop: 2i1 + 3(i1 + I) + i1 − V = 0 ⇒ i1 =
Right loop: E + 3(i1 + I) + 4I = 0 ⇒ E = − V2 −
2Ω
I1
V
3Ω
−
1Ω
− I2 ,
11I
2
(=
I
I1 + I
+
V
6
−
E
I
+
4Ω
V
2
+
11I
2
if polarity is opposite to that in figure).