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SYSTEM MATRIX HELP!











































Solve the system using matrices (row operations)
−4x−3y−z =-12
−5x+6y+4 =-24
−3x−y−6z =−22
How many solutions are there to this system?


A. None
B. Exactly 1
C. Exactly 2
D. Exactly 3
E. Infinitely many
F. None of the above

Note: each solution to this system is an ordered triplet with three coordinates. Together, the three coordinates make one solution. Having exactly three solutions is not possible.

If there is one solution, give its coordinates in the answer spaces below.

If there are infinitely many solutions, enter z in the answer blank for z, enter a formula for y in terms of z in the answer blank for y and enter a formula for x in terms of z in the answer blank for x.

If there are no solutions, leave the answer blanks for x, y and z empty.

x=

y=

z=

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Mar 25, 2011
Solve for Three Unknowns Using Matrices
by: Staff



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Part II


|………….:………|
| -4 -3 -1 :-12….|
| -5 +6 +4 : -24…|
|-3 -1 -6 :-22…...|
|…………:………| Multiply Row 1 by -1/4

|………….:………|
| 1 3/4 1/4 : 3….|
| -5 +6 +4 : -24…|
|-3 -1 -6 :-22…...|
|…………:………| add 5 times the 1st row to the 2nd row


|………….:………|
| 1 3/4 1/4 : 3….|
| 0 39/4 21/4 : -9…|
|-3 -1 -6 :-22…...|
|…………:………| add 3 times the 1st row to the 3rd row



|………….:………|
| 1 3/4 1/4 : 3….|
| 0 39/4 21/4 : -9…|
| 0 5/4 -21/4 :-13…...|
|…………:………| multiply the 2nd row by 4/39



|………….:………|
| 1 3/4 1/4 : 3….|
| 0 1 7/13 : -12/13…|
| 0 5/4 -21/4 :-13…...|
|…………:………| add -5/4 times the 2nd row to the 3rd row



|………….:………|
| 1 3/4 1/4 : 3….|
| 0 1 7/13 : -12/13…|
| 0 0 -77/13 :-154/13…...|
|…………:………| multiply the 3rd row by -13/77


|………….:………|
| 1 3/4 1/4 : 3….|
| 0 1 7/13 : -12/13…|
| 0 0 1 : 2…...|
|…………:………| add -7/13 times the 3rd row to the 2nd row



|………….:………|
| 1 3/4 1/4 : 3….|
| 0 1 0 : -2…|
| 0 0 1 : 2…...|
|…………:………| add -1/4 times the 3rd row to the 1st row


|………….:………|
| 1 3/4 0 : 5/2….|
| 0 1 0 : -2…|
| 0 0 1 : 2…...|
|…………:………| add -3/4 times the 2nd row to the 1st row


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

|………….:………|
| 1 0 0 : 4….|
| 0 1 0 : -2…|
| 0 0 1 : 2…...|
|…………:………|


The solution to the three simultaneous equations is:

From row 1, x = 4
From row 2, y = -2
From row 3, z = 2

The final answer to your question is:

x = 4
y = -2
z = 2


Check the answer by substituting the numerical values of x, y and z into the original equations:

-4x - 3y − z = -12

-4*4 - 3*(-2) - 2 = -12, correct


-5x + 6y + 4z = -24

-5*4 + 6*(-2) + 4*2 = -24, correct


-3x - y - 6z = -22

-3*4 - (-2) - 6*2 = -22, correct



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




Thanks for writing.


Staff
www.solving-math-problems.com



Mar 25, 2011
Solve for Three Unknowns Using Matrices
by: Staff


Part I
The question:
olve the system using matrices (row operations)
−4x−3y−z =-12
−5x+6y+4z =-24
−3x−y−6z =−22
How many solutions are there to this system?


A. None
B. Exactly 1
C. Exactly 2
D. Exactly 3
E. Infinitely many
F. None of the above

Note: each solution to this system is an ordered triplet with three coordinates. Together, the three coordinates make one solution. Having exactly three solutions is not possible.

If there is one solution, give its coordinates in the answer spaces below.

If there are infinitely many solutions, enter z in the answer blank for z, enter a formula for y in terms of z in the answer blank for y and enter a formula for x in terms of z in the answer blank for x.

If there are no solutions, leave the answer blanks for x, y and z empty.

x=

y=

z=

The answer:

First, arrange all three equations in standard form.

Ax + By + Cz = D

This has already been done in the statement of your problem:

−4x − 3y − z = -12

−5x + 6y + 4z = -24

−3x − y − 6z = −22

Prepare an Augmented Matrix of 3 rows and 4 columns.

Each row in the Augmented Matrix will represent one of the three 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, and the last number will be the constant:

|………….:………|
| -4 -3 -1 :-12….|
| -5 +6 +4 : -24…|
| -3 -1 -6 :-22…...|
|…………:………|

THIS IS OUR GOAL: Convert the Augmented Matrix into a triangular form. Once we accomplish this, we 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 1 0 : ?…|
| 0 0 1 : ?…...|
|…………………|

Once we 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.


To convert the augmented matrix into a triangular matrix, we will 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.

|………….:………|
| -4 -3 -1 :-12….|
| -5 +6 +4 : -24…|
|-3 -1 -6 :-22…...|
|…………:………|


Note:

There are many calculations involved, which means there is a high probability of making errors in your arithmetic. For this reason I recommend using a calculator if that is possible.



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