Math 317: Linear Algebra
Practice Exam 3
Fall 2015
Name:
*Please be aware that this practice test is longer than the test you will see on December
4, 2015. Also, this test does not cover every possible topic that you are responsible for
on the exam. For a comprehensive list of all topics covered on the exam, please see the
exam topics document on the website.
 
3
1
4

1. Find the closest point to y = 
5 in the subspace of R spanned by the vectors
1
v1 = (3, 1, −1, 1) and v2 = (1, −1, 1, −1).
We recall that the point in V that is closest to y is given by p = projV (y).
To compute p we observe that we have an orthogonal basis for V since
v1 · v2 = 0. Thus
y · v1
y · v2
p = projV (y) = projv1 (y) + projv2 (y) =
v1 +
v2
2
kv1 k
kv2 k2
 
 
3
1



(3, 1, 5, 1) · (3, 1, −1, 1)  1  (3, 1, 5, 1) · (1, −1, 1, −1) −1

=
−1 +
1
k(3, 1, −1, 1)k2
k(1, −1, 1, −1)k2
1
−1
 
   
3
1
3
 6 −1 −1
6 
1
  +   =  .
=
12 −1 4  1   1 
1
−1
−1
Recall that this particular formula for the projection only holds if we have
an orthogonal basis for V .
2. Let W be the subspace of R3 given by W = {(x1 , x2 , x3 ) : x1 + 2x2 + 3x3 = 0}.
(a) Find an orthonormal basis {b1 , b2 } for W .
We start by finding a basis for W . Noting that we have a single linear
equations, we obtain x1 = −2x2 − 3x
and
3. With
 x3 as the free
 x2 
−2
−3


variables, we obtain a basis for W as  1  ,  0  . We can apply


0
1
the Gram-Schmidtprocedure
to create an orthgonoal basis for W . To

−2
start, we let v1 =  1 . v2 is then given by
0
1
Math 317: Linear Algebra
Practice Exam 3
Fall 2015
  
 

−2
−3
−3/5
(−3, 0, 1) · (−2, 1, 0)   
1 = −6/5
v2 =  0  −
k(−2, 1, 0)k2
0
1
1
 
−3

Note that we can scale v2 by a factor of 5 to obtain v2 = −6.
5
Note that {v1 , v2 } is an orthogonal basis for W . To construct an
orthronormal basis for W , we divide v1 and v2 by their respective
magnitudes to obtain:
 
 
−2
−3
1  
v2
1  
v1
1 , b2 =
−6
=√
=√
b1 =

kv1 k
kv2 k
5 0
70 5 


(b) Find a third unit vector b3 that is orthogonal to both b1 and b2 so that
B = {b1 , b2 , b3 } is an orthonormal basis for R3 .
To find a third unit vector b3 that is orthogonal to both b1 and b2
3
so that B = {b1 , b2 , b3 } is an orthonormal basis
for R , we note that
b1 and b2 made up a basis for N (A) if A = 1 2 3 . Thus a basis

1
 
 
2
 
3
for V ⊥ is given by a basis for R(A) which is given by b3 = √12 +22 +33 ,
noting that the problem asked for a unit vector.
(c) Let T : R3 → R3 be the linear transformation that reflects x ∈ R3 across W .
Write the matrix of T with respect to B.
By definition, we have that T (x) = projV (x) − projV ⊥ (x). Now, since
b1 , b2 ∈ V and b3 ∈ V ⊥ , we have that
T (b1 ) = projV (b1 ) − projV ⊥ (b1 ) = b1 = 1b1 + 0b2 + 0b3
T (b2 ) = projV (b2 ) − projV ⊥ (b2 ) = b2 = 0b1 + 1b2 + 0b3
T (b3 ) = projV (b3 ) − projV ⊥ (b3 ) = −b3 = 0b1 + 0b2 − 1b3 .


1 0 0
Thus, [T ]B = 0 1 0 .
0 0 −1
(d) Write the matrix of T with respect to the standard basis.
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Math 317: Linear Algebra
Practice Exam 3
Fall 2015
To find [T ]stand , we use the change of basis formula: [T ]stand = P [T ]B P −1 ,
where the columns of P are the vectors in the basis B = {b1 , b2 , b3 }.
Thus, we have that
√
−2/√ 5
=  11/ 5
0
√

−2/√ 5
 11/ 5
0

[T ]stand
√
−3/√70
−6/√ 70
5/ 70
√
−3/√70
−6/√ 70
5/ 70
√ 

1/√14 1 0 0
2/√14 0 1 0 
3/ 14 0 0 −1
√ −1
1/√14
2/√14 .
3/ 14
3. Let

0
0
C=
1
0
1
0
0
0
0
1
0
0

0
0
,
0
2

1
0
D=
0
0

3 5 −9
2 −2 1 
.
0 1 −2
0 0
2
Compute det C, det D and det CD.

0
0
det C = det 
1
0
1
0
0
0
0
1
0
0


0
1


0
0
= − det 
0
0
2
0

1
0
det D = det 
0
0
0
0
1
0
0
1
0
0



0
1



0
0
= −
− det 0
0
2
0
0
1
0
0
0
0
1
0

0

0

0
2
= (1)(1)(1)(2) = 2

3 5 −9
2 −2 1 
 = (1)(2)(1)(2) = 4
0 1 −2
0 0
2
det CD = det C det D = (2)(4) = 8
4. (a) Let E be an elementary matrix that corresponds to one of the three types of
elementary row operations. Prove that det E = det E T .
Proof : There are three cases to consider. Suppose that E corresponds
3
Math 317: Linear Algebra
Practice Exam 3
Fall 2015
 
I1
 .. 
.
 
 Ij 
.

to switching ith row of A with the jth row of A. Then E = 
 .. , where
I 
 i
.
 .. 
In
Ij denotes the jth row of the identity matrix, and E is the identity
matrix except that Ij has been interchanged with Ii . Of course, this
changes the position of the 1 with respect to the columns of E. Thus
E = E T , i.e. E is symmetric and hence det E = det E T . Similarly, if
E corresponds to multiplying a row of A by a scalar c, then E is the
identity matrix except that the ith row of E has been multiplied by
the same scalar c. So E is symmetric and once again, we have that
det E = det E T . The third case follows in a similar manner.
(b) Prove that for any square matrix A, det A = det AT .
Proof : There are two cases to consider. If A is singular, then so is
AT , and so det A = 0 = det AT . Suppose that A is nonsingular. Then
we can write A as the product of elementary matrices. That is A =
E1 E2 . . . Ek where Ei is an elementary matrix. We have that det A =
det E1 . . . Ek = det E1 det E2 . . . det Ek = det E1T det E2T . . . det EkT =
det EkT . . . det E1T = det EkT . . . E1T = det(E1 . . . Ek )T = det AT .
(c) Prove that if A has a column of 0 entries then det A = 0 without directly using
the fact that A is singular.
Proof : We have shown that if A has a row of 0 entries then det A = 0.
If A has a row of 0 entries, then AT has a column of 0 entries. Since
det A = det AT , we have that det AT = 0.
5. Let A and B be 4 × 4 matrices and suppose that det A = 5 and det B = 3. Compute
the following if possible.
(a) det AT
det A = det AT = 5
(b) det A−1
A is nonsingular since det A 6= 0, and so det A−1 =
1
det A
= 14 .
(c) det AB
4
Math 317: Linear Algebra
Practice Exam 3
Fall 2015
det AB = det A det B = (4)(3) = 12
(d) det(A + B)
We do not have enough information to compute det(A + B).
(e) det(2A)
Note that there are 4 rows of A and each row has been multiplied by
2. Using our linearity properties, we can pull a 2 out from each row to
obtain det(2A) = 2 ∗ 2 ∗ 2 ∗ 2 det(A) = 16 det(A) = 16(5) = 80
(f) det(−B)
Similar to the problem above, we have that det(−B) = (−1)(−1)(−1)(−1) det B =
det B = 3
6. Determine whether {t, t + 1, t + 2} forms a linearly independent subset of P1 , the
space of all polynomials of degree 1 or less.
Proof : We claim that {t, t + 1, t + 2} forms a linearly independent subset
of P1 , the space of all polynomials of degree 1 or less. To see why, let
c1 (t) + c2 (t + 1) + c3 (t + 2) = 0 for all t ∈ R. We show that c1 = c2 = c3 = 0.
When t = 0, this implies that c2 + 2c3 = 0. When t = −1, this implies
−c1 + c3 = 0. Now, this gives us that c1 = c3 and c2 = 2c3 . Substituting
back into the original equation, we get that 4c3 (t + 1) = 0. Since this
equation must hold true for all t ∈ R3 ., we have that c3 = 0 which in turn
implies that c1 = c2 = 0.
7. Find a QR factorization for A and use this QR factorization to solve Ax = b where


2 1
1 1 

A=
0 1  ,
1 −1


2
1

b=
1
−1
Disregard this problem. The arithmetic is much too tedious for a type of
problem that you can expect to see on the exam. Instead look at problem
5 on Homework 9. My apologies about that.
5