Expert answer:Mathematics Assignment -linear equation system

Expert answer:Two equations: Solve the linear equation system by matrix equation Solve the linear equation system by Gauss-Jordan method Answer to every question step by step. Just do it as simple as you can, and please show all the steps from calculated questions. 🙂 I attached pdf file of the instructions, which you can use Thanks!
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Mathematics Assignment:
Problem 1. Solve the linear equation system by matrix equation (not in Excel).
 2 x+ 3 y+ 10 z = 63
 2 x -3 y -5 z = -30
 -4 x+ 3 y+ 5 z = 26
Problem 2. Solve the linear equation system by Gauss-Jordan method (not in Excel).
 -2 x -3 y+ 10 z = 37
 2 x+ 3 y -5 z = -12
 2 x+ 9 y+ 5 z = 56
Mathematics and Statistics for Business I
Figure 21. Three systems
(i)
x1
x2
2
x1
x2
2
(ii)
x1
x2
2
x1
x2
1
(iii)
x1
x1
x2
x2
2
2
The two lines in system (i) intersect at the point (2,0). Thus {(2,0)} is the solution set to (i).
In system (ii) the two lines are parallel. Therefore, system (ii) is inconsistent and hence its
solution set is empty. The two equations in system (iii) both represent the same line. Any
point on that line will be a solution to the system.
In general, there are three possibilities: the lines intersect at a point, they are parallel, or both
equations represent the same line. The solution set contains either one, zero, or infinitely many
points. We say that a linear system is consistent if it has either one solution or infinitely many
solutions; a system is inconsistent if it has no solutions.
The situation is similar for m×n systems. An m×n system may or may not be consistent.
►THEOREM 1. A system of linear equations has either
a) no solution
b) exactly one solution
c) infinitely many solution.
11.3 The Gauss-Jordan Elimination Method
This method is named in honor of Karl Friedrich Gauss (1777-1855), who is one of the
greatest mathematicians of all time and is often referred to as the “prince of mathematics”,
and Wilhelm Jordan (1842-1899), a German engineer.
The Gauss-Jordan method involves the repeated use of three basic transformations on a
system. We shall call the following transformations elementary operations.
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Mathematics and Statistics for Business I
Elementary row operations are
1) Interchanging two rows.
2) Multiplying a row by a nonzero real number.
3) Replacing a row by its sum with a multiple of another row, e. g. replace an
equation by itself plus a multiple of another equation.
Historical note
The Gauss solution method is so named because Gauss described it in a paper detailing the
computations he made to determine the orbit of the asteroid Pallas. The parameters of the
orbit had to be determined by observations of the asteroid over a 6-year period from 1803 to
1809. These led to six linear equations in six unknowns with quite complicated coefficients.
Gauss showed how to solve these equations by systematically replacing them with a new
system in which only the first equation had all six unknowns, the second included five
unknowns, the third equation only four, and so on, until the sixth equation had but one. This
last equation could, of course, be easily solved; the remaining unknowns were then found by
back substitution.
The Jordan half of the Gauss-Jordan method is essentially a systematic technique of back
substitution. In this form it was described by Wilhelm Jordan (1842-1899), a German
professor of geodesy, in the third (1888) edition of his Handbook of Geodesy.
EXAMPLE 1. Gaussian elimination.
Consider the linear system
The idea is to keep the first equation and work on the last two. In doing that, we will try to
kill one of the unknowns and solve the other two. For example, if we keep the first and second
equation, and subtract the first one from the last one, we get the equivalent system
Next we keep the first and the last equation, and we subtract the first from the second. We get
the equivalent system
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Mathematics and Statistics for Business I
Now we focus on the second and the third equation. We repeat the same procedure. Try to
kill one of the two unknowns (y or z). Indeed, we keep the first and second equation, and we
add the second to the third after multiplying it by 3. We get
This obviously implies that z = -2. From the second equation, we get y = -2, and finally from
the first equation we get x = 4. Therefore the linear system has one solution
Going from the last equation to the first while solving for the unknowns is called back
solving.
Keep in mind that linear systems for which the matrix coefficient is upper-triangular are easy
to solve. This is particularly true, if the matrix is in echelon form. So the trick is to perform
elementary operations to transform the initial linear system into another one for which the
coefficient
matrix
is
in
echelon
form.
Using our knowledge about matrices, is there any way we can rewrite what we did above in
matrix form which will make our notation (or representation) easier? Indeed, consider the
augmented matrix (where these are coefficients and right-side)
The entries appearing to the left of the dashed vertical line are the coefficients of the variables
as they appear in the system. This part of the augmented matrix is called the coefficient
matrix of the system. The numbers to the right of the dashed vertical line are the constants
on the right-hand side of the system as they appear in the system.
Let us perform some elementary row operations on this matrix. Indeed, if we keep the first
and second row and subtract the first one from the last one we get
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Mathematics and Statistics for Business I
Next we keep the first and the last rows, and we subtract the first from the second. We get
Then we keep the first and second row, and we add the second to the third after multiplying
it by 3 to get
This is a triangular matrix which is not in echelon form. The linear system for which this
matrix is an augmented one is
As you can see we obtained the same system as before. In fact, we followed the same
elementary operations performed above. In every step the new matrix was exactly the
augmented matrix associated to the new system. This shows that instead of writing the
systems over and over again, it is easy to play with the elementary row operations and once
we obtain a triangular matrix, write the associated linear system and then solve it. This is
known as Gaussian elimination.
Let us summarize the procedure of Gaussian elimination.
1. Construct the augmented matrix for the system;
2. Use elementary row operations to transform the augmented matrix into a triangular one;
3. Write down the new linear system for which the triangular matrix is the associated
augmented matrix;
4. Solve the new system. You may need to assign some parametric values to some unknowns,
and then apply the method of back substitution to solve the new system.
►THEOREM 1. In Gauss-Jordan elimination, we use elementary row operations on the
augmented matrix of the system to transform it so that the final coefficient matrix has a form
called reduced row echelon form with the following properties:
1) Any rows of zeros (called zero rows) appear at the bottom.
2) The first nonzero entry of a nonzero row is 1 (called a leading 1).
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Mathematics and Statistics for Business I
3) The leading 1 of a nonzero row appears to the right of the leading 1 of any preceding
row.
4) All the other entries of a column containing a leading 1 are zero.
Gauss-Jordan transformation step
1) We divide the leading row by the leading element.
2) We use elementary row operations (based on the leading row) to transform other
elements in the leading column to zero.
EXAMPLE 2. Gauss-Jordan elimination
Consider the linear system
We write the coefficients and the right-side into the Gauss-Jordan table
x
1
1
1
y
1
-2
2
z
1
2
-1
Right-side
0
4
2
We choose a leading element, for example in the first row and in the first column
leading
row
leading element
x
1
1
1
y
1
-2
2
z
1
2
-1
Right-side
0
4
2
R1=R1
R2=R2-R1
R3=R3-R1
Then the first column is called as leading column and the first row is called leading row. Now
we do the Gauss-Jordan transformation, i. e. the leading element must be transformed to 1
and the rest of the values in the leading column must be zeros after transformation and for
transformation we use the leading row.
x
1
0
0
y
1
-3
1
z
1
1
-2
Right-side
0
4
2
After the first Gauss-Jordan transformation we should do the second and the third
transformation, because we have 3 equations. Now we choose the second leading element.
We cannot use the previous leading row and previous leading column, i. e. we cannot use the
first row and the first column. For simplify the work the best choice for the second leading
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Mathematics and Statistics for Business I
element is in the second row number 1 or in the third row number 1. Number 1 is the best
choice because we do not need to divide the leading row by the leading element. For instance
we take the leading element now from the third row and make a Gauss-Jordan transformation.
x
1
0
0
y
1
-3
1
z
1
1
-2
Right-side
0
4
2
x
1
0
0
y
0
0
1
z
3
-5
-2
Right-side
-2
10
2
and we get the new table
R1=R1-R3
R2=R2+3R3
R3=R3
Now we choose the third leading element. We cannot use the previous leading rows and
previous leading columns, it means we have only one choice for the leading row and it is the
second row and the new leading column is the third column, where we have the number –5.
x
1
0
0
y
0
0
1
z
3
-5
-2
Right-side
-2
10
2
x
1
0
0
y
0
0
1
z
0
1
0
Right-side
4
-2
-2
and after transformation we get
And if we rewrite the Gauss-Jordan table, we get the linear system,
1 x 0 y 0 z
4
0 x 0 y 1 z
2
0 x 1 y 0 z
2
and the answer is x = 4, and y = -2 and z= -2.
Summary
1) A linear system has an associated augmented matrix, having the coefficient matrix of
the system on the left of the partition and the column of constants on the right of the
partition.
2) The elementary row operations on a matrix are as follows:
(Row interchange) Interchange two rows;
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Mathematics and Statistics for Business I
(Row scaling) Multiplication of a row by a nonzero scalar;
(Row addition) Addition of a multiple of a row to a different row.
3) A matrix is in row-echelon form if:
a) All rows containing only zero entries are grouped together at the bottom of the
matrix.
b) The first nonzero element (the pivot) in any row appears in a column to the right
of the first nonzero element in any preceding row.
4) A matrix is in reduced row-echelon form if it is in row-echelon form and, in addition,
each pivot is 1 and is the only nonzero element in its column. Every matrix is row
equivalent to a unique matrix in reduced row-echelon form.
5) In the Gaussian method with back substitution, we solve a linear system by reducing
the augmented matrix so that the portion to the left of the partition is in row-echelon
form. The solution is then found by back substitution.
6) The Gauss-Jordan method is similar to the Gauss method, except that pivots are
adjusted to be 1 and zeros are created above as well as below the pivots.
EXAMPLE 3. Solve the system by Gauss-Jordan method
x2
x1
x3
x2
2 x1
4 x2
3×1
x2
x4
x3
0
x4
6
x3
2 x4
2 x3
2 x4
1
3
At the beginning we write out the Gauss-Jordan table
x1
0
1
2
3
x2
-1
1
4
1
x3
-1
1
1
-2
x4
1
1
-2
2
Right-side
0
6
-1
3
R1=R1
R2=R2
R3=R3-2R2
R4=R4-3R2
and we choose the leading element from the second row and from the first column. After the
Gauss-Jordan transformation we get
x1
0
1
0
0
x2
-1
1
2
-2
x3
-1
1
-1
-5
212
x4
1
1
-4
-1
Right-side
0
6
-13
-15
R1=R1
R2=R2-R1
R3=R3+4R1
R4=R4+R1
Mathematics and Statistics for Business I
Now we take the second leading element from the first row and from the fourth column and
after transformation we get
x1
x2
x3
x4 Right-side
0
1
0
0
-1
2
-2
-3
-1
2
-5
-6
0
6
-13
-15
1
0
0
0
R1=R1+R3
R2=R2-2R3
R3=R3/(-2)
R4=R4+3R3
Now we take the third leading element from the third row and from the second column and
after transformation we get
x1
0
1
0
0
x2
0
0
1
0
x3
2
-3
2,5
2
x4
1
0
0
0
Right-side
7
-7
6,5
5
R1=R1-2R4
R2=R2+3R4
R3=R3-3R4
R4=R4/2
Now we take the fourth leading element from the fourth row and from the third column and
after transformation we get
x1
0
1
0
0
x2
0
0
1
0
x3
0
0
0
1
x4
1
0
0
0
Right-side
2
2
-1
3
The answer is x1= 2, x2= -1, x3= 3, x4= 2.
11.4 Linear System in Vector Notation
Consider the system
a 11 x1
a12 x2
… a1n x n
b1
a 21 x1
a 22 x 2
… a 2 n x n
b2
a n1 x1
an 2 x2
… a mn x n
bm
Above we saw that the system can expressed as a matrix equation
A X = B,
where A is the coefficient matrix, X is the variables matrix and B is the matrix of constants.
To discuss the problems of solvability we introduce the augmented matrix A1, which can be
obtained from matrix A by adding to it the column B, i. e.
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Mathematics and Statistics for Business I
10.5 Matrix Equation
One use of matrix multiplication arises in connection with systems of linear equations. We
consider a system of linear equations, where there are n variables and n equations
a 11 x1
a12 x 2
… a1n x n
b1
a 21 x1
a 22 x 2
… a 2 n x n
b2
a n1 x1
an 2 x2
… a nn x n
bn
we will let A denote the coefficient matrix of this system,
A
a11 a12
a1n
a 21 a 22
a 2n
,
a n1 a n2
a nn
X denote the column of variables
x1
X
x2
xn
and B denote the column
b1
B
b2
.
bn
Now we can write the linear equation system in form
AX=B.
For instance, the system
2x
y 4z 1
x 7y z
3
x
2
2y z
would be written
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Mathematics and Statistics for Business I
2 -1 4
x
1
1 -7 1
y
3
-1 2 1
z
2
as a matrix equation.
EXAMPLE 1. Weight reduction.
Bob weighs 178 pounds. He wishes to lose weight trough a program of dieting and exercise.
After consulting Table 1, he sets up the exercise schedule in Table 2. How many calories will
be burnt up exercising each day if he follows this program.
Calories burned per hour
Weight in lb
Exercise Activity
152
161
170
178
Walking 2 mph
213
225
237
249
Running 5,5 mph
651
688
726
764
Bicycling 5,5 mph
304
321
338
356
Tennis (moderate)
420
441
468
492
Table 1. Calories burned per hour
Exercise Schedule
Day
Walking
Running
Bicycling
Tennis
Monday
1
0
1
0
Tuesday
0
0
0
2
0,4
0,5
0
0
0
0
0,5
2
0,4
0,5
0
0
Wednesday
Thursday
Friday
Table 2. Hours per day for each activity
The information pertaining to Bob is located in column for of Table 1. This information can
be represented by a matrix X. The information in Table 2 can be represented by a 5×4 matrix
A. To answer the question, we simply calculate AX
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Mathematics and Statistics for Business I
1
0
1
0
0
0
0
2
0,4 0,5
0
0
0
249
764
356
0 0,5 2
0,4 0,5
0
492
0
605
Monday
984
Tuesday
481,6
Wednesday
1162
Thursday
481,6
Friday
►THEOREM 1. A system AX=B of n linear equations and n unknowns has a unique solution
if and only if A is invertible.
When the inverse of a square matrix A is known, we can easily find the solutions to a system
of linear equations
AX=B,
if we multiply this matrix equation by A-1 on the left, we have
A-1AX=A-1B
and hence the solution is given by
X=A-1B,
x
X
y
z
1
b1
a 11 a 12
a 1n
a 21 a 22
a 2n
b2
a n1 a n2
a nn
bn
We use this approach to solve the system in the next example.
EXAMPLE 2. Write down the following set of linear equations using matrices and solve the
matrix equation.
2x 3y
1
5 x 6 y 11
.
Solution
2
3
x
1
5 -6
y
11
and the solution is represented in the form
x
2
3
y
5 -6
200
1
1
11
.
Mathematics and Statistics for Business I
Now we compute
x
2
y
5 -6
1
3
1
11
1 -6 -3
27 – 5 2
1
-3
1
11
2
1
27
11
T
-6 – 5
1
27
27
1
27
1
.
And we get that x = 1 and y = -1.
EXAMPLE 3. Solve the system by the matrix method
x
y
z
5
2x
y 3 z 13 .
x 2 y 4 z 10
The solution is then given by
x
X
A 1B
y
z
1
18
1
1 1
1
2 -1
3
13
4
10
-1
2
– 10 – 2
4
5
5 -1
13
3 -3 -3
10
– 11
– 10/(-18) – 2/(-18)
5
4/(-18)
5
5/(-18) – 1/(-18)
13
0 .
3/(-18) – 3/(-18) – 3/(-18)
10
3
– 11/(-18)
We got that x = 2, y = 0 and z = 3.
EXAMPLE 4. Solve the matrix equation XA=B, where
1 -1 2
X
x y z, A
5 0 5 , B
0 3 6 .
3 2 4
Solution. If
XA=B
we multiply this matrix equation by A-1 on the right, and we have
XAA-1=BA-1
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Mathematics and Statistics for Business I
and hence the solution is given by
X=BA-1.
1 -1 2
X
BA -1
x y z
0 3 6
5 0 5
3 2 4
10.6 Exercises
1) Solve the system by the matrix method
2x
y 1
3x 8 y
Answer:
27
.
x
-1
y
3
.
2) Solve the system by the matrix method,
2x
y 3z
6
2x
y
12
z
4x 5 y
z
3
x
Answer:
– 33/2
y
12 .
9
z
3) Solve the system by the matrix method
x1
4 x2
3 x3
x1
2 x2
2 x1
2 x2
12
12
3 x3
.
8
x1
4
Answer: x 2
x3
4
.
– 8/3
4) Given
202
1
3 – 2,4 3 .
Mathematics and Statistics for Business I
A
5 3
6 2
, B
3 2
2 4
4 -2
, C
-6
3
.
Solve each of the following matrix equations
a) AX+B=C
b) XA+B=C
c) AX+B=X
d) XA+C=X
Answers: a)
20 – 5
– 34 7
, c)
0 -2
-2 2
.
5) Let
1 2 1
A
0 1 2
1 3 2
If possible, find a matrix B such that
AB=A2+2A.
3 2 1
Answer: B
0 3 2 .
1 3 4
6) Solve the matrix equation XA=B, where
X
x y, A
Answer: x y
5 0
2 3
, B
3 2 .
1/3 2/3 .
7) Solve the matrix equation XA=B, where
0 -1
X
x y z, A
2 -3 4 , B
4
Answer: x y z
2
3 2 1.
1 -1
4,7 – 1,7
1,6 .
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Mathematics and Statistics for Business I
8) Math for freaks! Solve the system by the matrix method
x1
2 x2
x3
x4
2 x5
2 x1
x2
4 x3
x4
5 x5
8 x1
x2
3 x3
x4
x5
4 x1
2 x2
3 x3 8 x 4
2 x5
5 x1
3×2
4 x3
6 x5
7 x4
1
16
1
.
5
7
1
1
Answer: X
1 .
1
2
204
…
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