Exercises Chapter I.
Page 7
3. Find all solutions, if any, of the following systems of linear equations:
a. 2x1 + x2 − x3 = 1
x1 + x2 + x3 = 2
2 1 −1 1
The augmented matrix of this system is
. After a few row operations
1 1 1 2
1 0 −2 −1
we get the matrix
. This matrix tells us that
0 1 3
3
x1 = −1 + 2x3
x2 = 3 − 3x3 .
b. 2x1 + x2 − x3 = 1
x1 + x2 + x3 = 2
x1 − x2
=2


2 1 −1 1
1 2. Row operations applied
The augmented matrix of this system is 1 1
1 −1 0 2


1 0 0 7/5
to this matrix lead to the matrix 0 1 0 −3/5. Thus,
0 0 1 6/5
7
x1 = ,
5
3
x2 = − ,
5
x3 =
6
.
5
4. Find all numbers k such that the following system of equations has a solution and then
solve that system.
2x1 + x2 − x3 = 1
x1 + x2 + x3 = 2
3x1 + 2x2
=k


2 1 −1 1
The augmented matrix of this system equals 1 1 1 2, after a few row operations
3 2 0 k


1 1
1
1
we get the following matrix 0 −1 −3 −3 . Thus, our system has a solution if
0 0
0 k−3
and only if k = 3.
7 Matt’s, Dave’s, and Abe’s ages are not known but are related as follows: The sum of Dave’s
and Abe’s ages is 3 more than Matt’s. Matt’s age plus Abe’s age is 9 more than Dave’s.
If the sum of their ages is 35, how old is Abe?
Let m, d, and a represent the ages of the three boys. Their ages satisfy the following
system of equations:
a+d = m+3
a+m = d+9
a + d + m = 35


1 1 −1 3
The augmented matrix associated with this system is 1 −1 1 9 . After some
1 1
1 35
row operations we get a matrix which indicates that Abe is six, Dave is 13 and Matt is
16.
14. Graph the solution sets for each of the following systems of equations. Since there are
two unknowns, the solution set is the set consisting of all pairs of numbers (x1 , x2 ) that
satisfy the equations
a. x1 − 2x2 = 6
2x1 + x2 = 1
2x1 + x2 = 1
x1
x1 − 2x2 = 6
( 85 , −11
)
5
x2
The graph shows there is only one solution: x =
8
5
and y = − 11
.
5
b. 3x1 + 4x2 = 1
1.5
1.0
0.5
K
2
K
0
1
1
2
x
K
0.5
K
1.0
There are an infinite number of solutions.
Page 17
1 2
3
−1
6 −4
1. Let A =
and let B =
. Compute the following matrices:
−1 0 −2
−2 −3
0
a. 3A b. A − B
c. 2A + 3B
1 2
3
3 6
9
a. 3
=
−1 0 −2
−3 0 −6
1 2
3
−1
6 −4
2 −4
7
b.
−
=
−1 0 −2
−2 −3
0
1
3 −2


1 0 −1
2 2 1

4. Compute if possible the following matrix
4. Let A =
. Let B = −2 1
−3 6 1
6 0
3
2
2
products: AB, BA, A = AA, and B = BB.
4 2 9
a. AB =
−9 6 30
b. The product BA is not defined. A has two rows, while B has three columns.
c. A2 = AA is not defined. A matrix can be squared only if it is a square matrix.


−5 0 −4
d. B 2 =  20 1 18 
24 0
3
1 6
x1 0
6. Let A =
. Are there any numbers x1 and x2 such that if B =
, then
−1 2
1 x2
AB = BA?
6
6x2 − 6x1
AB − BA =
. Thus, we need x2 − x1 = 0 and x2 − x1 = −1.
−x1 + 1 + x2
−6
Since
this is not possible there are no numbers xi for which the two matrices commute.
k
1+k
8. Find all numbers k such that the matrix K =
satisfies the equation K 2 = I2 .
1 − k −k
It turns out that the matrix K satisfies K 2 = I2 for all possible choices of the constant
k.
2 −1
2
3 6
11. Let A =
. Let B =
. Do there exist matrices X or Y such that the
4
1
1 −1 0
equations AX = B or Y A = B have solutions?
Comparing the sizes of A and B, we see that if such matrices exist then X must be a
2 × 3 matrix, and there is no matrix Y such that the product
Y A will be a matrix with
x1 x 2 x 3
3 columns, since A has 2 columns. Let X =
.
x4 x5 x 6
2x− x4 − 2 2x2 − x5 − 3 2x3 − x6 − 6
Then AX −B =
. This system of 6 equations
4x1 + x4 − 1 4x2 + x5 + 1
4x3 + x6
and 6 unknowns has a unique solution, which is:
x1 = 21 ,
x2 = 13 ,
x3 = 1,
x4 = −1,
x5 = − 73 ,
x6 = −4.
Page 27
1. Which of the following matrices are
0 1
1 −2
0 0
1
a.
,
,
,
1 2
0
1
1 0
0
in row echelon form?
0
0

 
 

1 1 1 0
0 1 2 4
1 1 0 −6
1
b. 0 1 1 3 , 0 0 0 1 , 0 1 1
0 0 1 0
0 0 1 6
0 0 0
1
a. The second and fourth matrices are in row echelon form.
b. The first and third matrices are in row echelon form.
2. Which of the following matrices are
0 1
1 −2
0 0
1
a.
,
,
,
1 2
0
1
1 0
0

1
b. 0
0

1
c. 0
0
 
1 1 0
0 1
1 1 3 , 0 0
0 1 0
0 0
 
2 0 0 1
1
0 1 2 8 , 0
0 0 0 1
0
in reduced row echelon form?
0
0
 

2 4
1 1 0 −6
0 1 , 0 1 1
1
1 6
0 0 0
1

2 0 0 0
0 1 2 0
0 0 0 1
a. Only the fourth matrix is in reduced row echelon form.
b. None of these matrices is in reduced row echelon form.
c. The second matrix is in reduced row echelon form.
3. Find the reduced row echelon form to which each of the following matrices is row equivalent.
6 1
−1
2
a.
,
2 4
3 −6

1

b. 4
3

1

c. 2
3
a.

−2 0
0 6 ,
0 8

−5 1
1
4
−13 10

2 8 −1
0 4 −2
0 1
1
1 0
,
0 1

1
b. 0
0

1
c. 0
0

6
3
3
1 −2
0 0

0 0
1 0 ,
0 1
0 0
1 0
0 1


1 0 0
0 1 0
0 0 1

3/5
12/5
−4/5
6. For each of the following elementary row operations, write the 3 × 3 elementary row matrix
that corresponds to it.
a. 2R2 + R3

1
a. 0
0

0
b. 0
1

1
c. 0
0
b. R1 ↔ R3

0 0
1 0
2 1

0 1
1 0
0 0

0 0
−3 0
0 1
c. −3R2
1 −2
9. Let E =
. Let A be any 2 × 2 matrix. What column operations performed upon
0
1
A will produce the product matrix AE?
Minus two times column 1 added to column 2.
12. An n × n matrix D = [dij ] is said to be a diagonal matrix if dij = 0 whenever
i 6= j. That
−1 0
is, the only nonzero entries of D are on the main diagonal. Let D equal
. For
0 2
each of the following matrices, A, compute DA and AD.
4 0
2 0
2 1
a.
b.
c.
0 3
1 0
0 0
−4 0
−4 0
a. DA =
, AD =
0 6
0 6
−2 0
b. DA
,
−1 0
−2 0
AD =
2 0
−2 −1
c. DA =
,
0
0
−2 2
AD =
.
0 0
16. How many different 2 × 2 matrices in reduced row echelon form are there? Find at least
four.
There are an infinite number of distinct 2 × 2 matrices in reduced row echelon form.
(The entry in the 1, 2 position can be arbitrary if the second row is a zero row.)
Examples of such matrices are:
0 1
,
0 0
1 0
,
0 0
1 1
,
0 0
1 0
0 1
Page 36
1. For each of the following systems of equations, write the corresponding coefficient matrix
and augmented matrix, and then determine the solution sets.
a. 10x1 − x2 + x3 = 9
−x1 + 6x2
=4
x1 + 5x2 − 7x3 = 1
b. −3x2 + 7x3 − x1 = −1
x3 + x1 − x2 = 4
x1 − x2 − x3 = 5
c. x1 + x2 = 1
x2 + x3 = 2
x3 + x4 = 3
x4 + x5 = 4
−x1 + x5 = 5
In the answers below, the first matrix
augmented matrix.



10 −1
1
10 −1
1



6
0 ,
−1
6
0
a. −1
1
5 −7
1
5 −7
87
61
, x2 = 106
, x3 = 106
.
x1 = 49
53



−1 −3
7
−1
1 ,  1
b.  1 −1
1 −1 −1
1
7
11
x1 = 4 , x2 = − 4 , x3

1
0

c. 
0
0
−1
1
1
0
0
0
0
1
1
0
0
0
0
1
1
0

0
0

0
,
1
1

1
0

0

0
−1
This system has no solution.
1
1
0
0
0
is the coefficient matrix and the second is the

9
4
1

−3
7 −1
−1
1
4
−1 −1
5
1
= −2.
0
1
1
0
0
0
0
1
1
0
0
0
0
1
1

1
2

3
.
4
5
2. Assume each of the following matrices is the coefficient matrix for a system of homogeneous
equations. Write the system and find all solutions.


−2
7
7 5
4
6 −2
a.
b.
c.  4 −1
3 2
−3 2
1
3
2
a. 7x1 + 5x2 = 0
3x1 + 2x2 = 0,
b.
x1 = 0, x2 = 0.
4x1 + 6x2 − 2x3 = 0
− 3x1 + 2x0 + x3 = 0
c. −2x1 + 7x2 = 0
4x1 − x2 = 0
3x1 + 2x2 = 0
x1 =
5
x,
13 3
x2 =
1
x.
13 3
x1 = 0, x2 = 0.
4 Assume each of the following matrices is the augmented matrix of a system of linear equations. Write the system and then find all solutions.


1 −1 0
2
4 6 0
1 1
1
a.
b. 1
−3 0 1
0
1 1 −2
a
4x1 + 6x2 = 0
− 3x1
=1
b. x1 − x2
=2
x1 + x 2 + x3 = 1
x2 + x3 = −2
x1 = − 13 , x2 =
2
9
x1 = 3, x2 = 1, x3 = −3
16. Solve the equation AX − XA = B for X, where A, B, and X are 2 × 2 matrices. The
matrix A equals
1 2
−1 4
while the matrix B is assumed known but arbitrary.
x1 x2
x2 + 2x3
−2x1 − 3x2 + 2x4
Suppose X =
. Then B = AX − XA =
.
x3 x4
−x1 + 3x3 + x4
−x2 − 2x3
b1 b2
Setting B =
, we want a solution to the equations
b3 b4
x2 + 2x3
−2x1 − 3x2 + 2x4
−x1 + 3x3 + x4
−x2 − 2x3
=
=
=
=
b1
b2
b3
b4
If the numbers bi satisfy the equations:
b1 + b4 = 0
3b1 + b2 − 2b3 = 0 ,
then the matrix equation B = AX − XA has a solution, which is
x1 = −b3 + 3x3 + x4
x2 = b1 − 2x3 .
Page 48
1. Determine which of the following matrices are invertible and then calculate their inverses

−1
2 3
a.
d.  0
1 4
1
2 3 1 0
1
a. The matrix
is row equivalent to
1 4 0 1
0
1
4 −3
inverse matrix is
2
5 −1
−2 −1
1 −2 3
b.
c.
6
3
−1
0 4

4 −1
2
1
0
3
0
4/5 −3/5
. Thus, the
1 −1/5
2/5
−2 −1 1 0
1 1/2 0 1/6
b. The matrix
is row equivalent to the matrix
, which
6
3 0 1
0 0 1 1/3
means there is no inverse.
1 −2 3
c. The matrix
is not a square matrix, which means it does not have an
−1
0 4
inverse.




−1 4 −1 1 0 0
1 0 3 0 0 1
1 0 1 0 is row equivalent to the matrix 0 1 1/2 0 1/2 0,
d. The matrix  0 2
1 0
3 0 0 1
0 0 0 1 −2 1
which tells us that this matrix does not have an inverse.
5. A biologist has been studying two different species of plants. Let xk and yk represent
the population of each species in the kth year. Suppose she has observed the following
relationships between the populations:
xk+1 = 2xk + yk
yk+1 = 3yk + xk
a. Find a matrix A such that
2 1
A=
.
1 3
xk+1
x
=A k
yk+1
yk
b. Show that
c. Show that
xk+1
yk+1
xk+1
2 xk−1
=A
yk−1
yk+1
xk
xk−1
= A
=A A
yk
yk−1
2 xk−1
= A
yk−1
xk
k x0
=A
y0
yk
for k = 1, 2, . . . .
The formula is certainly true when k = 1. An induction argument takes care of the
remaining values of k. Assume the formula is true for k, then
x+ 1
xk
k x0
k+1 x0
=A
=A A
=A
.
yk + 1
yk
y0
y0
d. If the population sizes in the third year are known, how can the initial populations be
determined; i.e.; what population size at time zero leads to the known populations?
x0
x
The matrix A has an inverse. Thus, from A
= 3 , we have
y0
y3
x0
−3 x3
=A
.
y0
y3
3
e. Compute A3 and A−3 .
15 20
A =
,
20 35
3
−3
A
1
7 −4
=
.
3
25 −4
11. Let E be the m × m elementary row matrix that corresponds to multiplying row i by
c(c 6= 0). Find E −1 .
The matrix E has zeros off the main diagonal, and all ones on the main diagonal
except at the [i, i] position which contains the number c. The matrix E −1 is identical
to the matrix E except at the [i, i] position, where c is replaced with 1/c.
15. Let D be any 2 × 2 diagonal matrix. Give a necessary and sufficient condition on the
diagonal entries so that D has an inverse. Compute the inverse of any such matrix.
d1,1 0
A necessary and sufficient condition for a 2 × 2 diagonal matrix D =
to be
0 d2,2
invertible is that each diagonal entry di,i 6= 0. Given D its inverse is
#
"
1
0
.
D−1 = d1,1
1
0 d2,2
Page 51
2. For each of the following find a system of equations for which the given statement is correct:
a. The solution set equals {(1, −2, 0)}.
x1 = 1 ,
x2 = −2 ,
x3 = 0 .
b. The solution set equals {(x1 , x2 ) : x1 = 2}.
x1 = 2 .
c. The system has no solution.
x1 + x2 = 1 ,
x1 + x2 = 2 .
4. If A is a finite set of numbers, let av(A) denote the numerical average of the numbers in
A. Thus, if A = {1, 2, −4}, then av(A) = − 13 . Let A = {a, b, c} and B = {a, b, c, d}.
a. Show that, in general, it is not true that av(B) = (av(A) + d)/2.
Suppose that the two numbers are equal. Then we have
a+b+c
+d
a+b+c+d
3
=
which implies that
4
2
3(a + b + c + d) = 2(a + b + c) + 6d
a+b+c
= d.
3
That is, d must equal the average of the first three elements in B. So suppose
A = {1, 1, 1} and B = {1, 1, 1, 0}. Then av(A) = 3, av(B) = 3/4,
and
av(A)+0
2
= 3/2
b. Let A1 and A2 be two sets of numbers. Find a condition that will guarantee
av(A1 ) + av(A2 )
av(A1 ∪ A2 ) =
2
Suppose the above equation is true, and that A1 has n members, while A2 contains m
numbers. Then we must have
n · av(A1 ) + m · av(A2 )
av(A1 ) + av(A2 )
= avA1 ∪ A2 ) =
n+m
2
2n · av(A1 ) + 2m · av(A2 ) = (n + m)(av(A1 ) + av(A2 ))
(n − m)av(A1 ) = (n − m)av(A2 ) .
So, if both sets have the same number of elements the equation is valid, or if their
averages are equal, then the equation is again valid.
a b
e f
c d
−1
6. Suppose A =
is invertible with A =
. Let B =
; that is, B is
c d
g h
a b
f e
obtained from A by a row interchange. Show that B −1 =
.
h g
0 1
Let E =
. Then B = EA and hence
1 0
B −1 = (EA)−1 = A−1 E −1 = A−1 E .
Multiplying a matrix on the right by E interchanges columns 1 and 2.