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Linear Equations - 4 Variables

by Katy Hadrava
(Bemidji, MN)










































Solve the system of linear equations and check any solution algebraically. (If there is no solution, enter NO SOLUTION. If the system is dependent, set w = a and solve for x, y and z in terms of a. Do not use mixed numbers in your answer.)

x + y + z + w = 13
2x + 3y − w = −1
−3x + 4y + z + 2w = 10
x + 2y − z + w = 1
(x, y, z, w) = (_,_,_,_)

Comments for Linear Equations - 4 Variables

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Dec 13, 2011
Linear Equations - 4 Variables
by: Staff

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Part IV


This is the final triangular form of the Augmented Matrix (the row echelon form):

| +1 +0 +0 +0 : +2 |
| +0 +1 +0 +0 : +0 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |


The solution to the four simultaneous equations is:

From R 1, x = 2
From R 2, y = 0
From R 3, z = 6
From R 4, w = 5


The FINAL ANSWER is:

x = 2
y = 0
z = 6
w = 5


This answer can be verified by substituting the numerical values of x, y and z into the original equations:


x + y + z + w = 13

2 + 0 + 6 + 5 = 13, correct


2x + 3y + 0z - w = −1

2*2 + 3*0 + 0*6 - 5 = −1, correct



-3x + 4y + z + 2w = 10

-3*2 + 4*0 + 6 + 2*5 = 10, correct


x + 2y - z + w = 1

2 + 2*0 - 6 + 5 = 1, correct




Since the numerical values of x, y, z, and w work in all four of the original equations, the solutions are correct.




Thanks for writing.

Staff
www.solving-math-problems.com

Dec 13, 2011
Linear Equations - 4 Variables
by: Staff


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Part III

Step 1

Add -2 times the R1 to R2

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| -3 +4 +1 +2 : +10 |
| +1 +2 -1 +1 : +1 |


Step 2

Add 3 times R1 to R3

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +7 +4 +5 : +49 |
| +1 +2 -1 +1 : +1 |


Step 3

add -1 times R1 to R4

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +7 +4 +5 : +49 |
| +0 +1 -2 +0 : +12 |


Step 4

add -7 times R2 to R3

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +0 +18 +26 : +238 |
| +0 +1 -2 +0 : +12 |



Step 5

add -1 times R2 to R4

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +0 +18 +26 : +238 |
| +0 +0 +0 +3 : +15 |



Step 6

multiply R3 by 1/18

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +0 +1 +13/9 : +119/9 |
| +0 +0 +0 +3 : +15 |


Step 7

multiply R4 by 1/3


| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +0 +1 +13/9 : +119/9 |
| +0 +0 +0 +1 : +5 |



Step 8

add -13/9 times R4 to R3

| +1 +1 +1 +1 : +13 |
| +0 +1 -2 -3 : -27 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |


Step 9

add 3 times R4 to R2


| +1 +1 +1 +1 : +13 |
| +0 +1 -2 +0 : -12 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |


Step 10

add -1 times R4 to R1


| +1 +1 +1 +0 : +8 |
| +0 +1 -2 +0 : -12 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |


Step 11

add 2 times R3 to R2


| +1 +1 +1 +0 : +8 |
| +0 +1 +0 +0 : +0 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |



Step 12

add -1 times R3 to R1


| +1 +1 +0 +0 : +2 |
| +0 +1 +0 +0 : +0 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |


Step 13

add -1 times R2 to R1


| +1 +0 +0 +0 : +2 |
| +0 +1 +0 +0 : +0 |
| +0 +0 +1 +0 : +6 |
| +0 +0 +0 +1 : +5 |

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Dec 13, 2011
Linear Equations - 4 Variables
by: Staff


----------------------------------------------

Part II

for the 1st equation: x + y + z + w = 13

| +1 +1 +1 +1 : +13 |

for the 2nd equation: 2x + 3y + 0z - w = −1

| +2 +3 +0 -1 : -1 |

for the 3rd equation: -3x + 4y + z + 2w = 10

| -3 +4 +1 +2 : +10 |

for the 4th equation: x + 2y - z + w = 1

| +1 +2 -1 +1 : +1 |



The complete Matrix:


| +1 +1 +1 +1 : +13 |
| +2 +3 +0 -1 : -1 |
| -3 +4 +1 +2 : +10 |
| +1 +2 -1 +1 : +1 |




THIS IS the GOAL: Convert the Augmented Matrix into a triangular form. Once this is done, you are finished.

The triangular form will look like this: a diagonal pattern of 1’s with 0’s everywhere else but the last column.

| 1 0 0 0 : ? = x |
| 0 1 0 0 : ? = y |
| 0 0 1 0 : ? = z |
| 0 0 0 1 : ? = w |


Once you have converted the matrix into the triangular form, the ? shown in row 1 will be the solution value of x, the ? shown in row 2 will be the solution value of y, and the ? shown in row 3 will be the solution value of z, and the ? in row 4 will be the value of w.


To convert the augmented matrix into a triangular matrix, you can perform various row operations, one at a time.

To convert the convert the Augmented Matrix to a triangular form, the following operations can be used:

1. switch two of the rows
2. multiply any row by a number which is not zero
3. Replace any row with the result of adding that row to another row.



Note:

There are MANY CALCULATIONS involved, which means there is a high probability of making errors in arithmetic. For this reason I RECOMMEND USING A CALCULATOR.


| +1 +1 +1 +1 : +13 |
| +2 +3 +0 -1 : -1 |
| -3 +4 +1 +2 : +10 |
| +1 +2 -1 +1 : +1 |

----------------------------------------------

Dec 13, 2011
Linear Equations - 4 Variables
by: Staff


Part I

Question:

by Katy Hadrava
(Bemidji, MN)


Solve the system of linear equations and check any solution algebraically. (If there is no solution, enter NO SOLUTION. If the system is dependent, set w = a and solve for x, y and z in terms of a. Do not use mixed numbers in your answer.)

x + y + z + w = 13

2x + 3y - w = −1

-3x + 4y + z + 2w = 10

x + 2y - z + w = 1

(x, y, z, w) = (_,_,_,_)





Answer:

There are several ways of solving this system of equations.

The following method is illustrated below: (1) preparing an Augmented Matrix, and then (2) using row operations to convert the Augmented Matrix into a triangular form. Once the Augmented Matrix has been converted to triangular form, the solution to every variable can often be read directly from the matrix with no further work.




First, arrange all three equations in standard form.

Ax + By + Cz + Dw = E

Standard form for all four equations:

x + y + z + w = 13

2x + 3y + 0z - w = −1

-3x + 4y + z + 2w = 10

x + 2y - z + w = 1


Prepare an Augmented Matrix of 4 rows and 5 columns.

Each row in the Augmented Matrix will represent one of the four equations.

The first number in each row is the coefficient for the x variable, the second number in each row is the coefficient for the y variable, the third number in each row is the coefficient for the z variable, the fourth number in each row is the coefficient for the w variable, and the last number will be the constant:

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Feb 24, 2017
Addition error NEW
by: Donald K

I believe an error within part III, December 13, 2011, step 5 after adding -1 times R2 to R4.

R4 should equal 39 instead of 15

R2 = (-1)(+0), (-1)(-2), (-1)(-3) : (-1)(-27)
= 0 2 3 : 27
+ 0 -2 0 : +12

= 0 0 3 : "39"

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