MATH 323.503
Exam 1 Solutions
October 19, 2015
1.
Consider the system of equations
4x + 12y − z = 6
−2x − 6y + z = −4
8x + 24y − z = 10
(a) Write down the augmented matrix for this system of equations with the variables
ordered x, y, z.
(b) Carry out the row reduction of the augmented matrix in (a) and transform it into
reduced row echelon form. Show all of the steps you are taking.
(c) Determine the complete set of solutions of this system of equations.
Solution: (a) We read off the augmented matrix from the system of equations:


4 12 −1 6
 −2 −6 1 −14  .
8 24 −1 10
(b) We perform row

4 12 −1
 −2 −6 1
8 24 −1
operations:


1
3
6
(1)
−14  −→  −2 −6
10
8 24

1 3 − 14
(3)
−→  0 0 1
0 0 0


3
− 14
1
2
(2)
1 −14  −→  0
−1 10
0


3
1 3
2
(4)


−2 −→
0 0
0
0 0

3 − 41 32
0 12 −1 
0 1 −2

0 1
1 −2  .
0 0
A few notes about the row operations performed: (1) Divide row 1 by 4; (2) Replace
row 2 with its sum with 2× row 1, and replace row 3 with its sum with (−8)× row
1; (3) Multiply row 2 by 2, note that it is the same as row 3, so can replace row 3 by
all 0’s; (4) Replace row 1 by its sum with 14 × row 2.
(c) From the row reduced matrix in (b), we see that the system of equations is
equivalent to
x + 3y = 1,
z = −2,
and so the solution set is
(1 − 3y, y, −2) y ∈ R .
1
2.
Let


1 a b
A = 0 1 a ,
0 0 1
where a and b are constants. Find A−1 .
Solution: There are a couple ways to do this problem. One is trial and error, though
this can be tricky. More straightforward is find A−1 by the direct method, so we row
reduce the following matrix:




1 a b 1 0 0
1 0 b − a2 1 −a 0
 0 1 a 0 1 0  −→  0 1
0 1 0 
a
0 0 1 0 0 1
0 0
1
0 0 1


1 0 0 1 −a a2 − b
−a  .
−→  0 1 0 0 1
0 0 1 0 0
1
Thus
A−1
3.
Let


1 −a a2 − b
−a  .
= 0 1
0 0
1


1 2 0
B =  1 0 2x .
x 1 1
For what values of x is B singular?
Solution: We take the determinant of B by expanding along the top row and find
|B| = 1 · (−2x) − 2 · (1 − 2x2 ) = 4x2 − 2x − 2.
Since
4x2 − 2x − 2 = 2(2x2 − x − 1) = 2(2x + 1)(x − 1),
we see that |B| = 0 if x = − 12 or x = 1 .
4.
For each statement below, write down whether it is True or False.
(a) If x, y, and z are non-zero vectors in Rn and x · y = x · z, then y = z.
(b) Suppose that x and y are non-zero vectors in in Rn and that a is the vector
projection of x along y. If kxk ≤ kak, then x and y are parallel.
(c) If A and B are square nonsingular matrices, then (BA)−1 = B −1 A−1 .
(d) Suppose that A and B are square 4 × 4 matrices and that B is obtained from A
by interchanging two rows and then multiplying a third row by 2. Then |A| = −2|B|.
(e) Suppose that A and B are square 4 × 4 matrices and that B is obtained from A
by replacing one row by 3 times a different row. Then |B| = 0.
2
(f) Suppose that A is a 3 × 3 matrix and that each column of A sums to 0. Then
|A| = 0.
(g) For n × n matrices A and B, if Ax = 0 has a nontrivial solution, then (AB)x = 0
has a nontrivial solution.
Solution: Credit was given for the correct answers. The explanations were not necessary, but the ones below may answer some of your questions about these problems.
(a) False : It is straightforward to find a counterexample. For example in R2 , take
x = [1, 0], y = [0, 1], and z = [0, 2].
(b) True : It helps to draw a picture, but one notes that the projection a of x along
y forms one of the legs of a right triangle with x as the hypotenuse. Thus we must
have kak ≤ kxk in all situations. If we are given that kxk ≤ kak, then it must be
that kxk = kak. This can only happen if x and y are parallel.
(c) False : The correct formula would be (BA)−1 = A−1 B −1 .
(d) False : We are applying elementary row operations to A and obtaining B, but
the correct formula would be |B| = −2|A|.
(e) True : The action performed here is not an elementary row operation. In fact a
row is being replaced by 3 times a different row, but not added to itself. The effect
here is that B has a row that is 3 times one of its other rows. This will force |B| = 0.
(f) True : In this case the rank of A must be at most 2, so |A| = 0.
(g) True : Since Ax = 0 has a nontrivial solution, it must be that |A| = 0. Therefore
|AB| = |A| · |B| = 0 · |B| = 0. Therefore, AB is a singular matrix, so the system of
equations (AB)x = 0 must have a nontrivial solution.
5.
Consider the matrix

1 0
1 1
A=
0 1
1 −1
0
0
1
1

1
0
.
1
4
(a) Show that the vector a = [1, −1, 1, 4] is a linear combination of the first three
rows of A.
(b) What is the rank of A? Explain.
Solution: (a) We want to find scalars a, b, c so that
a[1, 0, 0, 1] + b[1, 1, 0, 0] + c[0, 1, 1, 1] = [1, −1, 1, 4].
This amounts to solving the system of equations,
a+b=1
b + c = −1
c=1
a+c=4
3
So it must be that c = 1, and the second equation implies that b = −2. Then the first
equation implies that a = 3. We then verify that the fourth equation is also satisfied:
a + c = 3 + 1 + 4. Thus all four equations are satisfied by (a, b, c) = (3, −2, 1) .
(b) Since [1, −1, 1, 4] is in the row space of the top three rows, A is row equivalent to


1 0 0 1
1 1 0 0


0 1 1 1 .
0 0 0 0
A few elementary row operations then

1
0

0
0
yield that this matrix is row equivalent to

1 0 0
1 1 1
.
0 1 2
0 0 0
This matrix has three pivots, and so rank(A) = 3 .
6.
Let A be a nonsingular n × n matrix, and suppose that AT = A−1 .
(a) Prove that |A| = ±1.
(b) Prove that for any row vector x ∈ Rn we have kxAk = kxk.
Solution: Answers here may vary, but here is one solution.
(a) Since AT = A−1 , we know that AAT = I. Therefore, |AAT | = 1. However, we
also know that |AT | = |A|, and so
1 = |AAT | = |A| · |AT | = |A|2 .
Taking square roots, we see that |A| = ±1.
(b) Suppose that x ∈ Rn is a row vector. The key fact to use is that
kxk2 = x · x = xxT .
Similarly, xA is the product of a 1 × n matrix with an n × n matrix, and xA is also
a row vector, so we have
kxAk2 = (xA) · (xA) = (xA)(xA)T .
We continue the calculation, noting that taking transposes reverses the order of the
product:
kxAk2 = xAAT xT = xAA−1 xT = xxT = x · x = kxk2 .
The second equality follows from the given information that AT = A−1 . Now since
kxAk2 = kxk2 and since lengths of vectors are always nonnegative, it follows that
kxAk = kxk.
4