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JOSH BAR

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Use Gauss-Jordan row reduction to solve the system of equations. (If the system has no solution, say that it is inconsistent by entering NONE for each answer. If the system is dependent, enter a general solution in terms of x.)
-7x - y + 3z = 8
7x + y + 7z = 2

x=
y=
z=

I am not too familiar with the Gauss Jordan when it is set up in this format, any help would be appreciated.

Also if a row is a multiple of another row, what format could we see as far as the zeros?
 


Some people might say that it's kinda silly to employ Gauss-Jordan Elimination to an augmented coefficient matrix for solving this system, but what the hey.

?
Code: 
    [                      |         ]
    [    -7    -1     3    |    8    ]
    [                      |         ]
    [     7     1     7    |    2    ]
    [                      |         ]

This is the augmented coefficient matrix. Agree?

Here are two row operations:

R1 + R2 --> R2

R1 ÷ (-7) --> R1

In other words, first add the two rows, and then replace row 2 with the result. (Row 1 remains unchanged.)

Next, divide each element in row 1 by -7.

The resulting matrix clearly shows that z = 1.

Replace z with 1 in the equation represented by row 1; this gives an equation in x and y. Solve it for y.

Then you have:

x = x

y = some expresson in terms of x

z = 1

I did this in my head; I hope it's correct. 8-)



MY EDITS: Five typos so far; if any more remain, screw 'em.
 
JOSH BAR said:
… Also if a row is a multiple of another row, what format could we see as far as the zeros?
Click to expand...


I don't think that this situation would reveal itself in "zeros". I think that you would "see" it by comparing corresponding coefficients.

(I'm not sure that I understand your question about this.)

 
Hello, JOSH BAR!

Use Gauss-Jordan row reduction to solve the system of equations.

. . \(\displaystyle \begin{array}{ccc}\text{-}7x - y + 3z &=& 8 \\ 7x + y + 7z &=& 2 \end{array}\)
Click to expand...

\(\displaystyle \text{We have: }\;\left[ \begin{array}{ccc|c} \text{-}7 & \text{-}1 & 3 & 8 \\ 7 & 1 & 7 & 2 \end{array}\right]\)


\(\displaystyle \begin{array}{c} \\ R_2+R_1 \end{array} \left[\begin{array}{ccc|c} \text{-}7 & \text{-}1 & 3 & 8 \\ 0 & 0 & 10 & 10 \end{array}\right]\)


. . \(\displaystyle \begin{array}{c} \text{-}1\!\cdot\!R_1 \\ \frac{1}{10}R_2 \end{array} \left[\begin{array}{ccc|c}7 & 1 & \text{-}3 & \text{-}8 \\ 0 & 0 & 1 & 1 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1 + 3R_2 \\ \\ \end{array} \left[\begin{array}{ccc|c}7 & 1 & 0 & -5 \\ 0 & 0 & 1 & 1 \end{array}\right]\)


\(\displaystyle \text{We have: }\;7x + y \:=\:\text{-}5\)
. . . \(\displaystyle \text{and: }\;z \:=\:1\)

\(\displaystyle \text{Then: }\;\begin{Bmatrix}x &=& x \\ y &=& \text{-}7x-5 \\ z &=& 1\end{Bmatrix}\)


\(\displaystyle \text{On the right, replace }x\text{ with a parameter }t.\)

. . \(\displaystyle \begin{Bmatrix}x&=&t \\ y &=& \text{-}7t-5 \\ z&=&1 \end{Bmatrix}\)
 
As far as the second question, this is the exact question:

In the matrix of a system of linear equations, suppose one of the rows is a multiple of another. What can you say about the row-reduced form of the matrix?

a.The row-reduced matrix has two columns of zeros.
b.The row-reduced matrix has a row of ones.
c.The row-reduced matrix has a row of zeros.
d.The row-reduced matrix has a column of zeros.
e.The row-reduced matrix has two rows of zero
 
JOSH BAR said:
As far as the second question, this is the exact question:

In the matrix of a system of linear equations, suppose one of the rows is a multiple of another. What can you say about the row-reduced form of the matrix?

a.The row-reduced matrix has two columns of zeros.
b.The row-reduced matrix has a row of ones.
c.The row-reduced matrix has a row of zeros.
d.The row-reduced matrix has a column of zeros.
e.The row-reduced matrix has two rows of zero
Click to expand...

Start a new thread with a new problem.

Which of the five - you think - is the correct answer? Why?

Please show us your work/thought, indicating excatly where you are stuck - so that we know where to begin to help you.
 


Thank you, Soroban, for finishing the reduction. I should not have stopped at 10z = 10.

(I think that I was too eager to leave the matrix behind and get back to equations.)

Does reduced form require a 1 in element [1,1]? I'm thinking that the form with 7 is the difference between ref and rref.

Also, is it necessary to use the parameter t? The instructions call for a solution in terms of x.

 
I believe that the answer is b, as far as the 1st problem, I am confused because since it is a 2x4, the 1s and 0s are not in the place that they should be in.
 
Thanks to all, the answer did not require in the parameter of t, but I appreciate for going the extra step.
 
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