Mat 211
Dr. Firoz
2-3: System of equations and matrices
Systems, Matrices, and Applications
Systems of Linear Equations
System of equation
(Has solution) Consistent
Dependent
For Example: Consider the system
Inconsistent (has no solution)
Independent
3 x  2 y  1
5 x  3 y  11
Solve it and see that it has a unique solution. The system is consistent and independent.
The system
x y 4
2x  2 y  3
Is inconsistent, it has no solution. Check the solution.
Now the system
x y 4
2x  2 y  8
has infinitely many solutions, it is consistent and dependent.
We can solve it by the following methods:
1.
2.
3.
4.
5.
6.
Algebraically Using Elimination
Algebraically Using Substitution
Using Graphical method
Augmented Matrices and Row Operations.
Matrix algebra (the A1 B form)
Cramer’s Rule (Determinants).
Definition: The following representation of two linear equations is called the system of
linear equations in two variables:
x1  2x2  3
2 x1  x2  1
1
Solution: A solution of the given system is an ordered set or list of numbers that satisfies
all the equations simultaneously when we put x1  a and x2  b . This solution is written
as ( x1 , x2 )  (a, b) . We have our solution in this case as ( x1 , x2 )  (1,1) .
If the given system has at least one solution, it is said to be consistent. When the system
has no solution, it is said to be inconsistent.
Examples
1. The system x1  2x2  3 , 2 x1  x2  1 has unique solution. Find the solution.
2. The system x1  x2  2  2 x3 ,  x3  3x2  3 has infinitely many solutions.. Write
the solution in terms of the third variable.
3. The system 2 x1  2 x2  2 , x1  x2  3 is inconsistent.
4. Solve the system 4 x1  3x2  10 , 3x1  2 x2  18
5. If the system 2 x1  kx2  5 , x1  3x2  7 is inconsistent find k.
6. Solve the system x1  2 x2  x3  4 ,  x1  3x2  2 x3  4 , 2 x1  4 x2  x3  1
7. Consider this 2 by 2 system: use all the methods discussed above to solve the
system:
3 x  2 y  1
5 x  3 y  11
Solution:
a)
Algebraically Using Elimination
3(3x  2 y  1)  9 x  6 y  3
 19x  19  x  1
2(5 x  3 y  11)  10 x  6 y  22
By substituting x  1 we get y  2
b)
Algebraically Using Substitution
3x  2 y  1  y  3 / 2 x  1/ 2
So we can substitute y  3 / 2 x  1/ 2 in other equation
5 x  3  3 / 2 x  1/ 2   11  x  1
c)
Using Graphical method
3
1
3 x  2 y  1  y   x 
2
3
5
11
5 x  3 y  11  y  x 
3
3
2
d)
Augmented Matrices and Row Operations.
3x  2 y  1
5 x  3 y  11
2/3
 1/ 3  1 2 / 3
3 2  1  1 2 / 3  1/ 3 1






11  0  19 / 3 38 / 3  0
1
5  3 11 5  3
The last row gives y  2 , and by backward substitute we get x  1
 1/ 3 
 2 
Answers: 1. (1, 1) 2. ( x1  3  5x3 / 3, x2  1  x3 / 3, x3  x3 ) 3. No solution 4. (-2, 6)
5. k = 6
6. (1, 1,1)
Example 2. Let’s see another example by using Augmented Matrices and Row
Operations. Solve for x, y and z
x  2 y  6  4z
x  2 y  4z  6


or
 x  13z  6  y
 x  y  13z  6
2 x  6 y  z  1
2 x  6 y  z  1


Augmented Matrix:
 1 2 4
 1 1 13

 2 6 1
6
6 
1
Multiply row 1 by –1 and add with row 2 and get your new row 2
 1 2 4
0 3 9

 2 6 1
6
0  ; R2  R2  1R1
1
Multiply 2 with row 1 and add with row 3 and get your new row 3
1 2 4
0 3 9

0 2 7
6
0  ; R3  R3  2R1
11
Multiply row 2 by 1/3 gives you following
1 2 4
0 1 3

0 2 7
6
1
0  ; R2  R2
3
11
Multiply by -2 with row 2 and add with row 3 and get your row 3
1 2 4
0 1 3

0 0 1
6
0  ; R3  R3  2R2
11
3
Now by backward substitution to solve for variables:
z  11,
y  3(11)  0
y  33
x  2(6)  4(11)  6
x  104
 2 1 1
Now practice: 1 3 2

1 1 1
2
1 
2 
 Write the system of equations and solve.
Ans (2, -1, 1)
Reduced Row Echelon Form (rref):
The Elimination Method for solving large systems of linear equations
1. Make the leading coefficient 1 either by interchanging row or by multiplying or
dividing the first by a suitable constant.
2. Eliminate the leading coefficient each later equation by replacing the later
equation by the sum itself and a suitable multiple of the first equation
3. Repeat step 1 and 2for the 2nd row to eliminate leading coefficient 0 and to make
2nd element 1
4. Continue the process until all diagonal elements are 1 then do back substitution to
solve variables.
Example 3. Below are three row-reduced echelon forms for matrices of certain linear
systems. For each matrix, tell how many solutions the system has. Explain. If there are
any, find the solutions. If there are infinitely many solutions, find the general formula
and two particular solutions.
a.
b.
c.
[
[
[
]
]
]
1 0 0 3
0 1 0 − 2 The system has unique solution with x=3, y=-2, z=5
0 0 1 5
1 0 −2 0
0 1 4 0
0 0 0 1
The system does not have solution
1 0 −2 0
0 1 3 0
0 0 0 0
The system has many solutions, it is dependent and consistent.
 y  3z  0
x  2z  0
The general solutions are 
and 
x  2z
 y  3z
4
Now let’s discuss system of equations again:
1. Matrix algebra (the A1 B form)
Example: let
9
3 x  9  x   31 (9)  3
3
For matrix algebra we also use this inverse concept to solve systems of equation.
Singular Matrix: Singular matrices do not have an inverse. A matrix is singular
matrix if determinant of the matrix is equal to zero, let A is a matrix then A1 exist if
A 0.
So let’s discuss how to find determinant of a matrix
Finding determinant
If
b 
d b
a
1
A
then A  ad  bc and A1 

d 
ad  bc c a
c
2 
3
Example: A  
 then A  (9  10)  19  0
5  3 
Therefore A is not singular matrix. It has inverse. Find the inverse. And verify that
A1 A  AA1  I . What is I ? Answer: I is an identity matrix.
Finding Determinant of a 3 by 3 matrix
 1 2 4 
A   1 1 13 
 2 6 1 
| A | 1
1 13
1 13
1 1
 (2)
4
6 1
2  1
2 6
 1(1  78)  2(1  (26))  4(6  (2))  3
Note: The determinant of a matrix will be zero if
1. An entire row is zero.
2. Two rows or columns are equal.
3. A row or column is a constant multiple of another row or column.
Remember, that a matrix is invertible, non-singular, if and only if the determinant is not
zero. So, if the determinant is zero, the matrix is singular and does not have an inverse.
5
Sarrus Rule. In order to compute
Note: Sarrus rule is only applicable if the determinant is of order 3 by 3.
2 3 4
Example 1. Use Sarrus rule to find the value of A  4 3 1
1 2 4
12 + 4 + 48 = 64
2 3 4
4 3 1
1 2 4
2
4
1
3
3
2
 A  59  64  5
24 + 3 + 32 = 59
8 3 4
Exercise 1. Use Sarrus rule to evaluate the determinant A  4 6 1 and verify your
5 2 4
answer using calculator.
Finding inverse of a matrix:
Now let’s learn how to find inverse of a matrix. There are different methods to find
inverse matrix.
Method 1. Use Shortcut for 2 by 2 matrix
b 
a
1 d
Let A  
then A1 

d 
ad  bc  c
c
Example:
b 
a 
2 
 3
3
1
A
then A1 

3(3)  5(2)  5
5  3 
2
3

 2  19 19 


3   5 3 
19 19 
6
Method 2 (Optional). Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].
1
Example: Consider a matrix A  
3
calculator)
2 
and write the following (use rref using
4 
Method 3 (Optional). Adjoint method
A-1 =
(adjoint of A) or A-1 =
(cofactor matrix of A)T
2 3 4 
10  4  9 
1 


1
Let A  4 3 1 , then A  5 and A 
15 4
14 



5
1 2 4 
5
 1  6 
Now we know how to find inverse, let’s go back to solution of system of equations:
3x  2 y  1
Example 1. (continued) Given system is 
5 x  3 y  11
If we write in matrix form then we get the following,
2 
3
A

5  3 
x 
X  
 y
 1
B 
11 
If we do AX  B , we get the given system and we can rewrite
AX  B
 X  A1B
7
2 
 3
3
1
We have seen that A  
then A1 

3(3)  5(2)  5
5  3 
2
3
 2  19 19 


3   5 3 
19 19 
2
3

x
  19 19   1  1 
Now    
       so x  1 and y  2
 y   5 3  11   2 
19 19 
Cramer’s Rule (Determinants)
Given the system
This system has the unique solution
Example 1. (Continued) Solve (using Cramer’s Rule)
3 x  2 y  1
5 x  3 y  11
3 2
 (9  10)  19
5 3
3 1
Dy 
 (33  5)  38
5
11
D
Now, x 
Dx 19

1
D 19
and
Dx 
1
11
y
Dy
D
2
 (3  22)  19
3

38
 2
19
8
Example 2. Use Cramer’s Rule to solve the system:
4x - y + z = -5
2x + 2y + 3z = 10
5x – 2y + 6z = 1
Solution. We begin by setting up four determinants:
:
D consists of the coefficients of x, y, and z from the three equations
is obtained by replacing the x-coefficients in the first column of D with the constants
from the right sides of the equations.
is obtained by replacing the y-coefficients in the second column of D with the
constants from the right sides of the equations.
is obtained by replacing the z-coefficients in the third column of D with the constants
from the right sides of the equations.
Next, we evaluate the four determinants:
= 4(12 – (-6)) + 1(12 – 15) + 1(-4 – 10)
= 4(18) + 1(-3) + 1(-14)
= 72 – 3 – 14
= 55
9
= -5(12 – (-6)) + 1(60 – 3) + 1(-20 – 2)
= -5(18)+1(57) + 1(-22)
= -90 + 57 – 22
= -55
= 4(60 – 3) + 5(12 – 15) + 1(2 – 50)
= 4(57) + 5(-3) + 1(-48)
= 228 - 15 – 48
= 165
= 4(2 – (-20)) + 1(2 – 50) – 5(-4 – 10)
= 4(22) + 1(-48) – 5(-14)
= 88 – 48 + 70
= 110
Substitute these four values into the formula from Cramer’s Rule:
So, the solution is (-1, 3, 2).
10
Matrix Algebra on the Calculator TI:
1. Hit 2nd-MATRIX, then EDIT, and select [A], [B], etc.
2. Type in the size(order) of the matrix and the entries. Remember to hit ENTER after
each entry.
3.Repeat these steps until you have all your matrices entered.
4. To calculate hit 2nd-QUIT, then hit 2nd-MATRIX, then simply select
[A]. Then type the x-1 key, then hit ENTER. To convert this to fractions, hit
MATH, then select “Frac”, and hit ENTER.
5. To calculate A-1B, do as in step 4, but select [B] also, so that you have
[A]-1[B] on your screen. Hit ENTER to get the answer.
CASIO:
1. Hit MENU, and select 3:MAT. Select the matrix, type in its size and the
entries until you have all your matrices finished.
2. To calculate A-1, hit MENU, and select 1:RUN. Hit OPTN, then select
F2:MAT, then F1:MAT. This will bring up a “Mat” on the screen. Then you
hit ALPHA (Red key) and A (just below) to bring up A. Then hit the x-1 key
above the “)”.
3. Repeat these steps for A1 B .
Section: Gaussian Elimination
In this section we will learn a general method for finding possible solutions to a linear
system of equations. The method involves systematic elimination of the unknown from
each equation in turn. We will explain the method with examples.
Example 1. Solve the system
x1  x 2  3x3  5
2 x 2  x3  1
 3x1  2 x 2  2 x3  1
Solution: Now applying the operation R3  r3  3r1 we have the following
x1  x 2  3x3  5
2 x 2  x3  1
5 x 2  11x3  16
Applying R2  1 / 2 r2 we have
x1  x 2  3x3  5
x 2  .5 x3  .5
5 x 2  11x3  16
And by R3  r3  5r2
x1  x 2  3x3  5
x 2  .5 x3  .5
13.5 x3  13.5
11
Finally we the following by applying R3  r2 / 13.5
x1  x 2  3x3  5
x 2  .5 x3  .5
x3  1
We now have that x3  1 , and other unknowns can easily be found by backward
substitution into second and first equations. We have the solution ( x1 , x2 , x3 )  (1, 1, 1) .
This method is called the Gaussian Elimination method.
Augmented matrix: Let us consider the system of equations
x1  x 2  3x3  5
2 x 2  x3  1
 3x1  2 x 2  2 x3  1
The augmented matrix of the above system is
5
 1 1 3


 0 2  1 1
 3 2 2
1 

Row Echelon Form: A matrix is in row echelon form when
1. The entry in row 1, column 1 is a 1 and 0 appears below it
2. The first nonzero entry in each row after the first row is a 1, zeros appear
below it and it appears to the right of the first nonzero entry in any row above.
3. Any row that contains all zeros to the left of the vertical bar appear at the
bottom.
The row echelon form of the above augmented matrix is
5
1 1 3


.5  which exactly our Gaussian Elimination.
 0 1  .5
0 0 1
1 

Reduced Row Echelon Form (rref): Sometimes it is advantageous to write a matrix in
reduced row echelon form. In this form, row operations are used to obtain entries that
are zero above as well as below the leading 1 in a row. From reduced row echelon form
one can find solution of a system directly without backward substitution. The reduced
row echelon form of the above row echelon form will look like the one given below:
1
1 0 0


1 . The left hand side of the vertical bar is the identity matrix.
0 1 0
0 0 1
1

12
Finding reduced row echelon form by calculator:
TI:
1. Hit 2nd x 1 , then edit your matrix, hit 2nd MODE to go to the normal screen.
Hit 2nd x 1 then go to MATH, hit 8 for rref (and input your matrix A using
again 2nd x 1 , hit enter.
3. You will get your reduced row echelon form.
2.
CASIO:
1. Go to MENU, choose MAT, use right arrow to select size and edit your matrix
2. Go to OPTION and Press F1 for row operation
Example 2: Find the value of K so that the given system has no solution:
x1  x 2  x3  6
2 x1  x 2  x3  3
x1  2 x 2  Kx3  0
Solution: Find the determinant value of the coefficient matrix and set equal to zero.
1 1 1
2 1 1  0
1 2 K
Using Sarrus’s rule see that  K  3  (2 K  3)  0  K  2 . You may also use
row/column operation to find two zeros and find your result from reduced system.
x  4 y  1
Example: Use Cramer’s Rule to explain why the system 
has no solution.
2 x  8 y  3
Section: Matrices and Matrix Operations
Matrix: A matrix is simply a rectangular array of numbers considered as an entity. A
matrix with m rows and n columns in the array is written as follows:
 a11 a12  a1n 


 a 21 a 22  a 2 n 
A



 


a

a

a
m2
mn 
 m1
 b11 
 
b 
and a column vector looks like b   21  which is of order m  1

 
b 
 m1 
13
3 7 
  5  2
5 6 
 , B  
 and C  
 . Find the following
Example 1. Given A  
 5 12 
 3  4
 5 10 
results:
a) A  2B  C
b) A  B  0.5C
Example 2. Solve for x and y
0
0
3 2 2   x  y 2 2 6

1 0 1   4

x 6  5 2 x  y 5

 
Example 3.
 4 10 
1 3 2 
5




A   2  3
B  2 3 0 
C   1
6
0 1 5
 4
9 
A is a matrix 3 by 2, B is 3 by 3, C is 3 by 2
A  B  Not Possible
0
3 
7 
10  0  9
10 
4  5



A  C   2  (1)  3  3    3 0 
 6  4
9  7   10
16 
10  0   1 10
4  5

A  C   2  (1)  3  3    1  6 
 6  4
9  7   2
2 
Example 4.
 4 10 
A   2  3
6
9 
6
D  1 
0 
1 3 2 
B   2 3 0 
0 1 5
E  1 3
 4
5
C   1
 4
0
3 
7 
 1 2 
F
7 
8
2
G
1
5
6 
Order of above matrices:
A is a matrix 3 by 2, B is 3 by 3, C is 3 by 2, D is 3 by 1, E is 1 by 3, F and G are 2 by 2
matrices.
14
Find the following if possible
1. -B,
2.
3A-2C,
3. F+3G,
Answers:
 1  3  2 
 4 10 
5
 B   2  3 0 


3A-2C  3  2  3  2  1
0  1  5
6
 4
9 
 1 2   2
F  3G  
3
7  1
8
5  5

6 11
17 
25
4. 2B-5C
0  2
3    4
7  10
30 
 15
13 
2B-5C is not possible
Group work: now try
2  x 
 1 p 2 
 t 3
, B
, C
Consider A  
Find 2A + 2C, 2A+B



6 y 
 5 3  r
 s 2
bc 
1 a

Example 6. Given the matrix A   2 2b 2a  2c  . Find the determinant value of 4 A .
 3 3c 3a  3b 


Answer is zero.
Section: Matrix Multiplication
The multiplication of two matrices A and B is a unique matrix AB which is defined for A
of order m n , while B is of order n  p . Note that the order of AB is m  p .
3 7 
  5  2
5 6 
 , B  
 and C  
 . Find the following
Example 1. Given A  
 5 12 
 3  4
 5 10 
results:
a) BC
b) AB
c) BC  CA
Example 2. Find x
 4 5 4   x 
 x 2 1  5 0 1   1  O
 4 35 3   0 

 
x
 
   4 x  6 5 x  35 4 x  5   1  O
0
 
 x 4 x  6   5 x  35  3(4 x  5)   O
 4 x 2  6 x  5 x  35  12 x  15  0
Now solve for x (Answer should be 4, -5/4)
15
Example 3. Multiply the matrices
 a 1 a  2 6

 
1  a  a   2 2
3    2a  2(1  a) 6a  2(1  a)

0    2(1  a)  2a 6(1  a)  2a
3a 

3(1  a) 
2  1
 1 4 2 
 4 3
Example 4. Let A  
, B
, C


.
6 3 
 5 3  2
  5 2
following or explain why the answer does not exist.
c)
  7 5 6

Product AB. Answer: AB  
 9 33 6 
 16 7 

Sum 3A + 2C. Answer: 3 A  2C  
  3 12 
Product BC.
Answer: Does not exist
d)
Sum A + 3B.
a)
b)
Determine the
Answer: Does not exist
Matrix Multiplication:
Matrix Multiplication is possible if number of column of first matrix is equal to number
of row of 2nd matrix.
Example :
 4 10 
A   2  3
6
9 
1 3 2 
B   2 3 0 
0 1 5
Find the following:
AB  Not possible
5
C   1
 4
0
3 
7 
Because Ais 3 2 and B is 3  3
1(4)  3(2)  2(6)
BA   2(4)  3(2)  0(6)
0(4)  1(2)  5(6)
1(10)  3( 3)  2(9)
2(10)  3(3)  0(9)
0(10)  1( 3)  5(9)




10
  2
28
19
11
42




Now try for BC and CB .
16
Properties of addition and scalar multiplication:
( A  B)  C  A  ( B  C )
A B  B  A
A0  A
A  ( A)  0
(   ) A   A   A
 ( A  B)   A   B
( AB )C  A( BC )
A( B  C )  AB  BC
( A  B )C  AC  BC
AI n  I n A  A
where I n is n by n identitity matrix
Exercise 5. From the given matrices find ( I  A) D , where I is the identity matrix.
 0.2 0.5 
 200 
A
, D  

 0.3 0.4 
 400 
 1 1
2
3
4
n
Example 6. Let A  
 . Calculate A , A , A , then look at the pattern and find A ,
0
1


1 n
for n, a positive integer.
Answer: An  

0 1
Section 7.5 Rules for Matrix Multiplication (More examples Page # 282)
Example 1. Suppose P, and Q are n  n square matrices such that PQ  Q 2 P . Prove that
( PQ)2  Q6 P 2 .
Example 2. For the following matrices verify that AB  BA .
3 1
 0 3
A
B
,

6 2
 2 0
1 4
2
1
Example 3. A  
 , show that det(3 A)  3 det( A) , find also det((3 A) )
0 1
a b
2
Example 4. Prove that if A  
 , then A  (a  d ) A  (ad  bc) I , where I is a 2 by
c d 
2 identity matrix.
17
Section: The Inverse of a Matrix
For a given non-singular matrix A, the inverse matrix B  A1 exists such that
AB  BA  I , where I is an identity matrix of order same as A or B.
A matrix A is non-singular iff det( A) | A | 0
To find inverse of a nonsingular matrix using calculator:
Step 1. Input the matrix say A
Step 2. Call matrix A and hit x 1 in your calculator then hit MATH and select 1 :
to get the matrix along with determinant value.
Frac
Example 1. Find the inverse of a two by two matrix by hand:
d


d

b
a
b



1 
ad  bc
1
A
 then A 


c
ad  bc  c a  
c d 

 ad  bc
1
1
Now verify that AA  A A  I
b 
ad  bc 

a


ad  bc 

Example 2. Show that ( AB)1  B 1 A1
Solution: We can consider that
AB( AB) 1  I
A1 AB( AB) 1  A1 I  A1 , multiplying by A1
B 1 IB( AB) 1  B 1 A1 , using A1 A  I and multiplying by B 1
B 1 B( AB) 1  B 1 A1
( AB) 1  B 1 A1
Example 3. Show that ( ABC )1  C 1B 1 A1 using ( AB)1  B 1 A1
Example 4. Solve the following system of equations by matrix inverse:
 x  y  z  12

2 x  y  2 z  3
x  2 y  z  6

Solution: We have the following matrix system
 1 1 1 x   12 
 1 1 1

   


 2 1 2  y    3  , check for det  2 1 2   4
 1 2 1 z   6 
 1 2 1

   


 27 
1
 x   1 1 1  12 
 3 1 1 12   4 

  
   1
  
y

2

1
2

3


4
0
4

3


6


  
  

 
 z   1 2 1  6  4  5 1 3  6   45 
  
  

    
 4 
18
Example 5. Show that the following system has no solution using inverse matrix method.
 x  y  z  12

2 x  y  2 z  3
x  2 y  z  6

Section: Determinant of order 2 by 2 and Cramer’s Rule
We have discussed about it in the previous lecture. Check your note and also you may go
over your test book. Some more exercise problems are given below:
 a b
 c d
Exercise 1. Let A  
, B  
 show that | AB || BA |
 b a 
 d c 
1/ 2 1/ 2 
1 1 
2
2
2
2
Exercise 2. Let A  
, B  
 , calculate | A |  | A | , | B |  | B |
1
0
1

1




 .2 .6 .2 
Exercise 3. Let A   0 .2 .4  find | A  2I |1  | ( A  2 I )1 |
 .2 .2 0 


5a
2
1
Exercise 4. Show that
2
1
5a
2
Exercise 5. If 2 1  a
1
1
0  0 find all values of a. Answer: a  0, 1, 6
1 a
0
 2 x2
Exercise 6. If A  
 4x
1 a
0  a(1  a)(a  6)
0 1 a
3 x  1
 dA 
 determine det  
x 1 
 dx 
Answer: 4x 12
More examples:
Determinant:
2 by 2:
2 3
 18  15  3
5 9
a 2 1
3 by 3: 2 a 3  a
a 3
2
2 3
1
2 a
 a(a 2  6)  2(2a  3)  1(4  a)
2 a
1 a
1 2
1 2 a
Another way of expansion (making two zeros by operation)
19
a 2 1
2 a 3
1 2 a
R1 R1  aR3
R2  R2  2 R3

0 2  2a 1  a 2
2
1 a
2  2a 1  a 2
0 a  4 3  2a  1
 (1  a)
a  4 3  2a
a  4 3  2a
1
2
a
 (1  a)(6  4a  (a 2  3a  4))  (1  a)(10  a  a 2 )  ( a  1)( a 2  a  10)
Now verify the following:
1.
7 17 29
a) 11 19 31  36
13 23 37
2. Prove the following:
a
b
a)
b
c
265 240 219
b) 240 225 198  0
219 198 181
3
c) 2
4
1
0
3  28
1 4 4
ax  by
bx  cy  (ax 2  2bxy  cy 2 )(b 2  ac)
ax  by bx  cy
0
yz
zx 0
1 x
b) 1 y
x y
1 z
1
c)
2
3
a a b  a(b  3a)
a 0 b
1
1
2
2
d) a
b
b3
a3
a
1
c 2  (a  b)(b  c)(c  a)(ab  bc  ca)
c3
b
c
e)
a
b
c 2  (a  b)(b  c)(c  a)(a  b  c)
bc ca ab
f)
a  bc b  ca c  ab  (a  b)(b  c)(c  a)(a 2  b 2  c 2 )
a3
b3
c3
2
2
1
1
2
bc
a
g)
b
ca
c
( a  b) 2
ca
h)
bc
1
2
c
2
a
b
 4abc
ab
ca
(b  c) 2
ab
bc
ab  2abc(a  b  c)3
(c  a ) 2
20
abc
i)
a
b
b  c  2a
b
 2(a  b  c)3
a
c  a  2b
c
c
3. Solve for all x, a, b, or c:
1 1 1
a
a) x a b  0
b) a 2
x
2
a
2
b
2
Answer: a) x  a  b
b
b2
c
c2  0
bc ca ab
b) a  b  c  0
1
2
3
a b  0
a 0 b
c) a
c) a  b  0
Matrices:
5 3
4. Given A  
, find A2

 2 4
1 
2
5. Given A    , B  3 2 4 5 find BA
3
 
4
5 3
5 2 
6. For a given matrix A  
, the transpose is A  

 , first row becomes
 2 4
3 4 
first column and second row becomes second column. A matrix A is called
orthogonal (perpendicular) if AA  AA  I . Show that the matrix
1 2 2
1
A   2 2 2 is orthogonal.
3
 2 2 1
9 1
1 5 
, B
7. Given A  

 . If 3A  5B  2 X  O , where O is the zero
 4 1
7 15
matrix, find the matrix X.
8. If  4 5   x y   7 3 , find x and y
2 0 1
9. Suppose that A   2 1 3 , and f ( x)  x 2  5 x  6 find f ( A)


1 1 0 
 4
 4 8 4


10. Given 1   x y z    1 2 1  , find x, y and z
 3
 3 6 3
11. Solve the following system using both Cramer’s rule and matrix inverse:
21
2 x  3 y  3
a) 
4 x  y  11
2 x  3 y  z  12  0

d) 3x  4 y  11z  46
5 y  4 z  5

5a  6b  4c  15

b) 7a  4b  3c  19
2a  b  6c  46

2 x  3 y  z  110

c) 3x  5 y  2 z  190
3z  y  x  120

2 x  y  5
e) 
3x  2 y  3
 240 750 
1200 1500 
 210 
12. Given an input matrix A  
 and the demand matrix D  
.
450 
330 
 720
1200 1500 
Find the output matrix X, such that ( I  A) X  D
2 x  3 y  3
13. Find k so the system has no solution: 
kx  y  11
2 x  3 y  3
14. How do you determine whether the system 
has infinitely many
 x  1.5 y  11
solutions or no solution at all?
2 x  3 y  3
15. How do you determine whether the system 
has infinitely many
 x  1.5 y  1.5
solutions or no solution at all?
16. Find an equilibrium vector for the transition matrix given below:
0.4 0.4 0.2
P   0.3 0.3 0.4
0.4 0.4 0.2
Solution: The equilibrium vector is a row matrix v  [ x
that vP  v . Use matrix multiplication and find v.
0.4 0.4 0.2 
[ x y z ]  0.3 0.3 0.4   [ x y z ] 
0.4 0.4 0.2 
y z], x  y  z  1 such
0.4 x  0.3 y  0.4 z  x
0.4 x  0.3 y  0.4 z  y

Solve this system 
0.2 x  0.4 y  0.2 z  z
 x  y  z  1
4
4
3
x  0.36364  , y  0.36364  , z  0.27272 
11
11
11
4 4 3
The required equilibrium vector is v  
which is a 1 by 3 matrix.
11 11 11 
22