Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
MATH 1180: 
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
Solutions to Assignment #4
1. (a) It is known that matrix multiplication is not commutative, but is associative. Given the following
 1 2 
 0 1
3 4
 2 3
 , Q  
 , R  
 , and S  
 ;
matrices: P  
 1 3
 1 2 
2 0
 4 1
Verify that: (PQ)RS = P(QR)S.
1
 PQ  
 1
2  0

3    1
 2
( PQ) R  
 3
1   02

2   0  3
3  3

7   2
1 4 
 
1  6 
4   66
 
0    9  14
 2

 3
80 
 
 12  0 
3 

7 
0

5
8
 12
08 
 0  8    2 3   0  32
 
  
 
 ( PQ) RS  
5

12
4
1

10

48
15

12


 

0 0 
 0 1  3 4   0  2
2
 
  

Now, QR  
 
40 
 1 2  2 0    3  4
1
Pre-multiplying QR by P gives:
 1 2   2 0   2  2 0  8   0  8 
 
  
  

P(QR )  
 1 3  1  4   2  3 0  12   5  12 
And finally, post-multiplying P(QR) by S gives:
0  8    32
 0  8   2 3   0  32

  
  
P(QR ) S  
 5  12  4 1    10  48 15  12    58



  32

  58
0

 4 
8 

 3 
8 
  ( PQ) RS …..verified.
 3 
(b) By applying the associative property of matrix multiplication on the product B –1 A–1(AB), or
otherwise, prove that: (AB) –1 = B –1 A–1.
 From the associative property, then:
B –1 A–1(AB)
= B –1 (A–1A)B
= B –1 (I)B
= B –1 B
=I
So, the product (B –1 A–1) (AB) equals the identity matrix I.
But the inverse of matrix (AB) is unique, with: (AB) –1 (AB) = I
 (AB) –1 = B –1 A–1. ■
(c) Using the identity in part (b) above, prove that: (ABC) –1 = C–1 B–1 A–1 .

(ABC) –1 = C–1 (AB)–1
= C–1 (B–1 A–1)
= C–1 B–1 A–1 ■
b
a
 , where a, b, c, and d have integral values, such
2. (a) Find the four (4) matrices of the form A  
d 
c
that the matrices have a determinant of 1 and are also orthogonal.
 Firstly, the determinant of A is given by: |A| = ad – bc = 1
Secondly, the orthogonality of A is defined as: AT = A–1 .
c
 b 1  d
 b  d
a
1  d

  

 , and A 1 
Now: AT  
a  1  c
a    c
A  c
d
b
a
So then, equating AT and A–1 gives: 
b
c  d

d    c
 b

a 
 b

a 
1
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
Equating corresponding matrix elements on the L.H.S. and R.H.S., implies:
a=d
…..(1)
c = –b
…..(2)
b = –c
…..(3)
d=a
…..(4)
Equations (1) and (4) are equivalent, while (2) and (3) are equivalent.
So, collectively we have: d = a and c = –b.
a
Hence, at this stage, the matrix A  
c
And |A| = 1
b
 a
 is refined to become: A  
d
 b
b
.
a 
 a2 – (b)(–b) = 1

a2 + b 2 = 1
…..(5)
But a and b are integers, so equation (5) means these two variables can only take possible
values of 0, 1, or –1. Any other integer values outside –1, 0, and 1, simply violates Eqn.(5).
So, let’s start assigning (or substituting) values to fit equation (5):
*a = 0  b = 1.
 0
(two matrices)  A1  
1
That is: a = 0, b = 1; a = 0, b = –1
 1

0 
1
0
 ; A2  
0
1
*a = 1  b = 0.
1
That is: a = 1, b = 0 (one matrix)  A3  
0
0

1
*a = –1  b = 0.
1
That is: a = –1, b = 0 (one matrix)  A4  
 0
0

 1
Now, for values of b:
*b = 0  a = 1.
 1
(two matrices)  A5  
 0
That is: b = 0, a = 1; b = 0, a = –1
0
 1
 ; A6  
1
 0
0

 1
And, observe matrices A5 and A6 are identical to A3 and A4 , respectively.
 0
Next: *b = 1  a = 0. This would give: A7  
1
0
And, finally: *b = –1  a = 0, gives: A8  
1
1
 …….. identical to matrix A1 .
0 
 1
 …….. identical to matrix A2 .
0 
So collectively, we have only four (4) unique matrices.
These are:
 0
A1  
 1
1
0
 ; A2  
0
1
 1
1
 ; A3  
0
0
0
1
 and A4  
1
 0
0
 . Ans.
 1
 x 1
2
 such that AT + |A|A–1 = 
(b) Find, if possible, all matrices of the form A  
 y z
0
y
 b   z  1
x
 d
 , and |A|A–1 = adjoint of (A)  
=

 Firstly: AT  
z
a    y
x 
1
 c
0
.
2 
 1
y  1
 x z

 = 
z  x 
x
1  y
So equating our obtained algebraic-matrix expression for AT + |A|A–1 with the given numerical matrix:
y  1  2 0 
 x z

= 
 ,  x + z =2, y – 1 = 0, 1 – y = 0, and z + x = 2.
z  x   0 2 
1  y
Therefore:
AT + |A|A–1 =
x

1
y
+
z 
 z

 y
2
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
The first & third equations are equivalent, and both imply (generally) that: z = 2 – x.
The second & third equations are also equivalent, and imply: y = 1.
So with the above inter-variable (x, y, and z) relationships,
 x 1
x
 is refined to become: A  
the given matrix A  
 y z
1
1 

2  x 
Ans.
3. (a) Prove that:
(i)
(AB)T = BTAT …….. from first principles.
 Let
A = [aij] of order mn
B = [bij] of order np
Then C = AB = [cij]
of order mp
n
The element in the ith row and jth column of AB is given by cij =
a
k 1
ik
bkj p
And this is also the element standing in the jth row and ith column of (AB)T.
Now, let’s obtain a similar-type expression for the elements in the product BTAT :
The elements of the jth row of BT are: b1j , b2j , b3j , . . . , bnj
{there are n-columns in BT}
(‘formerly’, the elements of the jth row of B)
And, the elements of the ith column of AT are: ai1 , ai2 , ai3 , . . . , ain
{there are n-rows in AT)
(‘formerly’, the elements of the jth row of B)
n
Therefore, the element in the jth row and ith column of BTAT is:
b
k 1
n
But the product (of two numbers): bkj aik = aik bkj . Hence:
 bkj a ik =
k 1
Thus
(ii)
(AB)T
=
BTAT.
(A–1)T = (AT) –1
kj
a ik
n
a
k 1
ik
bkj = cij = (AB)T
■
 Hint: Consider the product AT(A–1)T .
 Obviously: AT (A–1)T = (A–1A)T
= IT
= I
But, AT (A–1)T = I  (A–1)T is the inverse of AT .
That is, (A–1)T = (AT) –1 .
(iii)
■
A3 = A  A = A–1.
 A3 = A2 A
Hence, the given expression A3 = A  A2A = A
 A2 = I.
 A A = I  A is the inverse of A.
That is: A = A–1.
■
(b) Prove the following statements (assuming square-, multipliable, and/or invertible- matrices):
(i) If A and B commute (that is, AB = BA), and A and B are both symmetric, then AB is symmetric.
 So we are required to prove that AB is symmetric, that is (AB)T = (AB).
First: (AB)T = BTAT
= B A  {A and B are symmetric, so BT = B and AT = A}
= (A B) ….. shown. ■
3
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
(ii) If A and B are orthogonal, then AB is orthogonal.
 We are basically required to prove (with the given conditions) that: (AB)T = (AB)–1 .
First: (AB)T = BTAT
= B–1A–1 {both A and B are orthogonal}
= (AB) –1 …… proven. ■
(iii) If A and B are symmetric, then AB – BA is skew-symmetric.
 Recall, that a matrix X is skew-symmetric if XT = –X.
Now, (AB – BA)T = (AB)T – (BA)T
= BTAT – ATBT
= BA – AB
= –(AB – BA)  AB – BA is skew-symmetric.
■
(iv) If A is an m-square symmetric matrix and P is of order m×n, then PTAP is symmetric.
 We are basically required to prove (with the given conditions) that (PTAP)T = PTAP .
First: (PTAP)T = [PT (AP)]T {simple associativity}
= (AP)T (PT)T
= (AP)T P
= (PTAT) P
=(PTA) P
{remember A is symmetric  AT = A}
= PTAP
….. proven.
■
* So we have shown that the transpose of the matrix PTAP is equal to the said matrix PTAP, hence
proving that PTAP is indeed symmetric.
(v) If A is an m-square skew-symmetric matrix and P is of order m×n, then PTAP is skew-symmetric.
 This time we are required to show (with the given conditions) that (PTAP)T = –(PTAP).
First: (PTAP)T = [PT (AP)]T {associativity}
= (AP)T (PT)T
= (AP)T P
= (PTAT) P
=[PT (–A)] P
{remember A is skew-symmetric  AT = –A}
= – PTAP
….. proven.
■
4. By appropriate successive pre-multiplication and/or post-multiplication of the given equations by a
matrix (or its inverse), obtain the following results from the given equations.
(i) Given that AB–1C = D, then C = BA–1D and A = DC–1B.
 Starting with what is given: AB–1C = D
Let’s solve for C first.
Then, clearing matrix A from the L.H.S. by pre-multiplying the given equation by A–1, gives:
(A–1 A)B–1C = A–1 D

B–1C = A–1 D
{recall A–1 A = I}
Then, clearing B–1 from the new L.H.S. by pre-multiplying through by B, gives:
(BB–1) C = BA–1 D

C = BA–1 D ■
4
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES

Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
For the second proof and all subsequent proofs in this overall question, we shall use as a general
notation “A” to mean pre-multiply through by matrix A, and also “A” to mean post-multiply
through by matrix A. Of course, too, it is already appreciated that A–1 A = I and A A–1 = I, and that
matrix products such as IC and CI (as a general example) returns the matrix C this will be done
implicitly (without writing IC or CI, as the case may be). Note also, that the associative rule of
matrix multiplication does allow A–1 A to be evaluated instantly in the context of say multiple items
such A–1 ABC to give simply BC, or PQBB–1 to give PQ .
So let’s carry out the second proof of part(i), to solve for A, as follows:
We are given:
C–1
B


AB–1C = D
A B–1 = D C–1
A
= D C–1 B
■
(ii) Given that A + B = AB, then A–1 + B–1 = I.

We are given:
A + B = AB
A–1   I + A–1 B = B
 B–1
 B–1 + A–1 = I
Re-arranging L.H.S.:  A–1 + B–1 = I
■
(iii) Given that A + B = BC, then A–1 + B–1 = CA–1.

We are given:
 A–1
A + B = BC

I + B A–1 = BCA–1
B–1  
B–1 + A–1 = CA–1
And, re-arranging L.H.S.:  A–1 + B–1 = C A–1
■
(iv) If A–1 + B = A + B–1, then: (A – B) = A(B – A)B.

We are given:
A–1 + B = A + B–1, {and we need to clear all inverses for the final answer.}
A
 I + AB = AA + A B–1
B
 B + ABB = AAB + A
Rearranging both sides:
 A – B = ABB – AAB
= A(BB – AB)
= A(B – A)B …… shown.
4
4
  10


1  and B   15  4
 5
4 
1

2

5. (a) (i) If A   4
1

3
2

 AB =  4
1

4
  10

 15  4
 5
1

3
3
2
4

1
4 
3
2
■
9

 14  , show that: AB = 5I.
6 
9    20  45  20 8  12  4
 
 14     40  45  5 16  12  1
6    10  30  20 4  8  4
18  42  24   5
 
36  42  6    0
9  28  24   0
0
5
0
0

0  = 5I. ■
5 
Hence, solve the system of equations: –10x + 4y + 9z = 10, 15x – 4y – 14z = 20, –5x + y + 6z = –30.
4
  10

 The system of equations in matrix form is:  15  4
 5
1

9   x   10 
  

 14   y    20  ,
6   z    30 
or in condensed matrix algebra format:
BX = C, where B is the coefficient matrix, while X and C
are the column matrices on the L.H.S. and R.H.S., respectively. It is therefore required to solve for the
column of independent variables, X.
Of course from matrix theory: BX = C  X = B–1C.
5
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
We need B–1, the inverse of B:
 ( 15 A) is the inverse of B.
Recall earlier, that AB = 5I. This implies that ( 15 A)B = I
Therefore, B–1 = 15 A. Hence our solution now takes the form: X =
x
2
  1
That is:  y    4
 z  5 1
 

3
3
2
1
5
AC.
4  10 
 20  60  120 
  40    8 

 1
 1
 

1  20    40  60  30    70    14 
5
 5   70    14 
4   30 
10  40  120 

 

Hence, the solution to the given system of equations is:
x = –8, y = 14, and z = –14.
■
(ii) Express matrix A in part(i) above as the sum of a symmetric and a skew-symmetric matrix.
 Hint: if A is a square matrix, then A + AT is symmetric, while A – AT is skew-symmetric.
A
= ½ (A + AT) + ½ (A – AT)
 2
1 
= p  4
2 
1

3
4
1
7
2
5
7
 A =
3
2
6
3
4  2
 
1   3
4   4
4
3
1
 2
1 

1 
2  +  4
2 
1
4 

1
5
 0

1
3 
 1
2
 3
8 

0
1
3

 1
0 
3
3
2
4  2
 
1   3
4   4
4
3
1
1 

2  p
4 
Ans.
* Please check (visually) that the two matrices are indeed symmetric and skew-symmetric.
Watch how the elements reflect about the leading diagonal mirror-line. The symmetric
matrix reflects onto the exact elements, while the skew-symmetric matrix reflects onto the
negative value of the elements. Also, it is automatic (by its definition) that a skew-symmetric
matrix has a leading diagonal comprising only of zeros. Unlike any skew-symmetric matrix,
however, there are no constraints on the leading diagonal of a symmetric matrix.
(b) Find the rank of each of the following three matrices: [Use row-by-row reduction, if necessary].
1 2 3 


A   2 3 4,
3 5 7 


1

2
B
4

2

1
4
1
4
6

 1 2 1 4 


1 6
,
C

2
4
3
5

.
2 9
 1  2 6  7 



2 7 
1
 Remember that the rank r of any matrix A is the order of the smallest square sub-matrix of A that
has a non-zero determinant. We also know for an nn square matrix (order = n), if its determinant is
zero, then its rank will be less than n. Row by row reduction is one way of obtaining the actual rank.
And moreover, any matrix having one row (or column) being a linear combination of others rows (or
columns) in the original matrix, will have a decrease in rank (and a determinant of zero, in the case of a
square matrix).
Let’s organize the given problem as parts (i), (ii), and (iii) for the matrices A, B, and C respectively.
(i)
matrix A :
Observe the third row of the square matrix A is the sum of the first and second rows. Hence its
determinant is zero, and its rank, r, is definitely less than its order n, where obviously n = 3.
Let’s carry out the row-by-row reduction:
1 2 3 
1 2 3 




R 3 (R1  R 2 )
P 2 3 4
 2 3 4 ,
3 5 7 
0 0 0




And, it is now clear, after the row-by-row reduction, that the smallest square sub-matrix having a nonzero determinant has an order two. Hence, rank A = 2.
 

6
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
Please observe for yourself that there a three possible 22 square sub-matrices (after row reduction of
2  2
3
1
1 3 
  and, incidentally, they all have non-zero
 , 
 , and 
the matrix A). These are: 
3  3
4
 2 4
2
determinants. However, it is not a requirement that all the sub-matrices have a non-zero determinant,
only at least one.
You may feel free to carry out the row reduction up to an upper-triangular form (for an original
square matrix, as A) as you are accustomed. We did the R1 – (R2 + R3) operation simply because
it was observed upfront that R1 is equal to R2 + R3 , as stated earlier. So our subsequent row
reduction was conveniently planned around this fact.
(ii)
matrix B :
We will just create zeros below the leading diagonal in a column-by-column manner, going from
left to right. A line is drawn through the leading diagonal just to show visually, that the aim of
the subsequent row-reductions is to continually create zeros are to be created below this line,
until the matrix become at least upper-triangular.
1

2
p
4

2

1
1
4
1
4
1
2
2
6

6
9

7 
  
R 4  2R1 ; R 3  4R1 ; R 2  2R1
  

R 4  R 2 ; 2R 3  3R 2
1

0
0

0

1
2
0
0
6 

1 6 
 7  24 

1
1 
1
1

2
0
0  3

0
2

6

 1  6
p
 2  3

0  5 
1
1


7R 4  R 3
1

0
0

0

1
2
0
0
6 

6 
….. FINAL.
 7  24 

0  17 
1
1
And, we see that the determinant of B is not zero, because we have no row (or column) in the
final row-reduced upper-triangular form of B that is comprised entirely of zeros.
Hence: rank B = 4.
(iii)
matrix C: { 34 matrix}
 1 2 1 4 


 2 4 3 5 
 1  2 6  7 


 
R 3  R1 ; R 2  2R1
 1

 0
 0

2 1
0
5
0
5
4 

3 
 3 


R 3 R 2
 1

 0
 0

2 1
0
5
0
0
4 

3 
0 
And, we see clearly that there no 33 sub-matrices with non-zero determinants, due to the
zeros in the entire third row of the row-reduced form of C. However, we do identify [at least
one] 22 sub-matrices having non-zero determinants. Hence, rank C = 2. ■
6. (a) Use Row-by-Row elimination to show that following system of equations has infinite solutions.
6x + 3y + 2z = 6
10x + 5y + 6z = 10
2x + y – 3z = 2 .
Obtain the solution to the system (in terms of one or more free variables, as appropriate).
 We will row-reduce the augmented matrix “A:H” of the given system until the A-portion of this
matrix becomes upper-triangular.
 6

 10
 2

3
2
5
6
1 3
6 

10 
2 


3R 3  R1 ; R 2 5R 3
 6

 0
 0

3
2
0
21
0  11
6 

0
0 
 

21R3  11R 2
 6

 0
 0

3
2
0
21
0
0
6 

0
0 
7
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
And, we have exactly one entire row of zeros, which means that the system has infinite solutions with
(correspondingly) exactly one free variable. Students should be observant enough to note that
although the middle matrix above was already in upper-triangular form, the third row could be further
reduced using the second row, and this was done (thereby creating an entire row of zeros).
* For those of us who like the more abstract ‘rank’ treatment; then, from the final row-reduced A:H
matrix, we see also that rank(A) = rank(A:H) = 2 < n, where n = 3 (independent variables).
And, of course, “rank(A) = rank(A:H) < n” implies, infinite solutions with n – r free variables, where r is
the rank of A. That is, 3 – 2 = 1 free variable.
 Now to solve the system (using the final row reduced A:H matrix):
 6

 0
 0

3
2
0
21
0
0
6 

0
0 
R2  21y = 0  y = 0
R1  6x + 3y + 2z = 6  6x + 0 + 2z = 6
 z = ½ (6 – 6x)
 z = 3 – 3x
Let x be the free variable, since we already have z in terms of x.
Hence, let x =  ….. some free (real number) parameter.  z = 3 – 3
Therefore, collectively, the solution to the system is: {x = , y = 0, z = 3 – 3}. ■
and
(b) In trying to solve a system of equations, the final row-reduced form of the augmented matrix was
as follows:
1

0
0

2
c
3
d
0
a
1 

e 
b 
Choose any set of values for the constants a, b, c, d, and e that would make the system:
(i)
Have unique solutions
 For unique solutions, we just make sure rank(A) = n = 3. Or put another way, we make sure
that A-portion of the A:H augmented matrix above has a non-zero determinant.
In that case, we may use, for example: a = 1, c = 1 ….. to keep the leading diagonal of A free of
zeros. This guarantees a non-zero determinant.
(ii)
Have infinite solutions with one free variable
 One free variable ‘means’ exactly one entire row of zeros. Hence: a = 0, b = 0.
(iii)
Have infinite solutions with two free variables
 Two free variables ‘mean’ exactly two entire rows of zeros.
Hence: a = 0, b = 0, c = 0, d=0, e= 0.
(iv)
Be inconsistent.
 Inconsistency (no solution) free variables means that at least one row in A is comprised
entirely of zeros, but the corresponding row in A:H is not comprised entirely of zeros. Or, more
visually, inconsistency occurs when an entire row in A is zero, but the neighboring element (to
the right) in the same row of the H column is not zero.
Hence, in the present context, we have: a  0, b = 0. And an example is: a = 1, b = 0.
■
8
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
7. Find the values of the rational numbers a and b that makes the following system
x – 2y + 3z = 4
2x – 3y + az = 5
3x – 3y + 5z = b .
has:
(i) unique solution,
(ii) infinite solutions, (iii) no solution.
Find the unique solution corresponding to the values a = 3 and b = 0.
Find the set of infinite solutions.
 Again, we start with the augmented matrix:
2
1

2
3

4

a
5 p
5
b 
1

R 3 3R 2
0
0

3
3
3
1  2

1
0
0
3

2
3
4 

1
a6
3 
0 14  3a b  3 


a6
3 
4
b  12 
3



R 3 3R1 ;R 2  2R1
4
So, we have our final row-reduced form of the augmented matrix, in upper-triangular form.
 (i) UNIQUE SOLUTIONS:
 rank(A) = n = 3.
 14 – 3a  0
 a  14/3 .
 (ii) INFINITE SOLUTIONS:  at least one entire row of zeros.
 14 – 3a = 0 and b – 3 = 0
 a = 14/3 and b = 3.
 (iii) NO SOLUTIONS:
 rank(A) < rank(A:H).
 14 – 3a = 0 and b – 3  0
 a = 14/3 and b  3.
 Unique solution corresponding to the values a = 3 and b = 0.
We substitute a = 0 and b = 3 into the final row-reduced A:H matrix as follows:
1

0
0

2
1
0
And:


a6
3 
14  3a b  3 
3
4

a  3, b  0
1

0
0

2
3
1
3
0
5
4 

3 
 3 
R3  5z = –3  z = –3/5
R2  y – 3z = –3  y = 3z – 3 = 3(–3/5) – 3
 y = –22/5
R1  x – 2y + 3z = 4 
x = 4 + 2y – 3z
= 4 + 2(–22/5) + 3(–3/5)
x = –33/5
Therefore: Required unique solution is: {x = –33/5, y = –22/5, z = –3/5}.
9
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
 Set of Infinite Solutions:
We substitute a = 14/3 and b = 3 into the final row-reduced A:H matrix as follows:
1

0
0

2
1
0


a6
3 
14  3a b  3 
3
4
R2  y –
And:
4
3

a  14/3, b 3
z = –3  y = –3 +
R1  x – 2y + 3z = 4 
4
3
2
1

0

0
z
3

1
0
4
3
0
4 

3 

0 
….let z be the free variable.
x = 4 + 2y – 3z
= 4 + 2(–3 +
4
3
z) – 3z
x = –2 – 13 z
So letting the free variable, z = , then collectively the set of infinite solutions is:
{x = –2 – 13 , y = –3 + 43 , z = }.
■
8. Find the values of the real scalars m and n for which the set of equations
x + 3y + 2z
(m–1)y + 2z
2x + my + (m–2)z
has:
= 9
= 2
= n–2 .
(i) unique solution,
(ii) infinite solutions, (iii) no solution (inconsistency).
Find the unique solution corresponding to the values m = 0 and n = 0.
Find the set of infinite solutions corresponding to the highest value of m.
 We start with the augmented matrix of the given system:
1

0
2



m 1 2
2 
m m  2 n  2 
3
2
9
 
(m 6)R 2 ; (m 1)R 3


R 3 R 2
1

0
0



1

0
0

R 3  2R1
1

0
0



m 1
2
2

m  6 m  6 n  20 
3
2
9


(m  6)(m  1)
2(m  6)
2(m  6)

(m  1)(m  6) (m  1)(m  6) (m  1)(n  20 ) 
3
2
9


(m  6)(m  1)
2(m  6)
2(m  6)
 ……FINAL.
0
(m  6)(m  3) (m  1)(n  20 )  2(m  6) 
3
2
9
Again, we have the final row-reduced augmented matrix, in upper-triangular form.
 (i) UNIQUE SOLUTIONS:
 rank(A) = n = 3.
 (m – 6) (m – 3)  0
 m  3 and m  6.
 (ii) INFINITE SOLUTIONS:  at least one entire row of zeros.
Observing the 2nd row of the augmented matrix, we see that m = 6 makes it entirely zeros.
But the last row (3rd row) isn’t entirely zeros, as yet due to the element in the lower righthand corner of the matrix. In fact substituting m = 6 makes the augmented matrix looks like:
10
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
1

0
0

3
2
0
0
0
0
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010


0
 m= 6
5(n  20 ) 
9
And infinite solutions are only guaranteed if in addition to m = 6, we have 5(n – 20) = 0, that is
the extra condition n = 20. Hence, one set of infinite solutions (with two free variables, due to
the two rows of zeros) will occur when: m= 6 and n = 20.
Also, observing the last row of the final row-reduced augmented matrix, we see that m = 3 may
give infinite solutions if some additional conditions on the element in the “3 rd row, 4th column”
position are imposed. Let’s see. By substituting m = 3 into the final A:H matrix, we get:
1

0
0

3
2
9
6
6
6
0
0
2n  34


 m = 3


And, an entire row of zeros is obtained only if 2n – 34 = 0. That is, if n = 17.
Hence, a second set of infinite solutions (with one free variable, due to only one row of zeros)
will occur when: m= 3 and n = 17.
 (iii) NO SOLUTIONS:
Observing the final row-reduced forms of A:H when m= 6 and m =3 were substituted, we see
that inconsistency (no solution) will occur when: 5(n – 20)  0 and 2n – 34  0, respectively.
So inconsistency is achieved when either:
{m = 6 and n 20} OR {m = 3 and n 17}
■
 Unique solution corresponding to the values m = 0 and n = 0.
Substituting m= 0 and n = 0 into the final row-reduced A:H matrix gives:
1

P 0
0

And:
3
2
6
 12
 12
0
18
32





9
R3  18z = 32  z = 16/9
R2  6y – 12z = – 12, or simplified: y – 2z = –2  y = 2z – 2
= 2(16/9) – 2
y = 14/9
R1  x + 3y +2z = 9  x = 9 – 3y – 2z
 x = 9 – 3(14/9) – 2(16/9)
 x = 7/9
So, collectively, when m= 0 and n = 0, the unique solutions is: { x = 7/9, y = 14/9, z = 16/9 }.
 The infinite solution corresponding to the highest value m .
We revisit our ‘infinite solution’ possibility-set obtained previously, in part (ii). And we
therefore use m = 6 (and n = 20) to get the A:H matrix to look like:
1

0
0

3
2
9
0
0
0
0
0
0


  m = 6, n = 20.


11
Course: Engineering Mathematics I (MATH 1180)
Solutions to Assignment #4: MATRICES
Group I: CIV, ELEC, SURV.
Lecturer: Mr. Oral Robertson
Date: November 4, 2010
And we observe two entire rows of zeros, to expect two free variables in our infinite solution.
Now, R1  x + 3y +2z = 9  x = 9 – 3y – 2z
We have x in terms of y and z, so let y and z be the two free variables.
Let y =  and z =  (real parameters) ; therefore: x = 9 – 3 – 2.
So collectively, the required infinite solution set is: { x = 9 – 3 – 2, y = , z =  }
■
9. Define the following terms:

(i)
column matrix  a matrix having only one column.
(ii)
square matrix  a matrix having equal numbers of rows and columns.
(iii)
singular matrix a matrix having a determinant of zero.
(iv)
adjoint of a [square] matrix A the product of the determinant of matrix A and its inverse.
(v)
rank of a matrix the order of its smallest square sub-matrix having a non-zero determinant.
(vi)
transpose of a matrix A the matrix obtained when the rows and columns of A are
interchanged.
(vii)
leading diagonal the line of elements in a square matrix that goes from the top left-hand
corner of the matrix to the bottom right-hand corner of the matrix.
(viii)
symmetric matrix  a square matrix that is equal to its transpose.
(ix)
skew-symmetric matrix a square matrix that is equal to the negative of its transpose.
(x)
upper-triangular matrix a square matrix where all elements below the leading diagonal are
zeros.
E N D O F S O L U T I O N S.
12