Question

matrix with gauß elimination.
I cant found Z2, because it's false.
My try is on picture.

Answer 1

whats the question here?

Answer 2

gauss elmination

Answer 3

I was never taught this sorry

Answer 4

source of exercise:

Answer 5

\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
j&-j&-j&0\\
1&1&1&6\end{array}\right)\]
First set of row operations: \(jR_1-R_2\to R_2\) and \(R_1-R_3\to R_3\) (where\(R_i\) denotes the \(i\)-th row, and \(\to\) indicates replacement)
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&j^2+j&0&2j+2j^2\\
0&j-1&-2&6j-4\end{array}\right)\]
Next, \(\dfrac{1}{j^2+j}R_2\to R_2\) gives
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&j-1&-2&6j-4\end{array}\right)\]
Then \(R_2+R_3\to R_3\):
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&j&-2&6j-2\end{array}\right)\]
Then \(jR_2-R_3\to R_3\):
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&0&2&-4j-2\end{array}\right)\]
Then \(\dfrac{1}{2} R_3\to R_3\):
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&0&1&-2j-1\end{array}\right)\]
Finally, \(R_1-jR_2\to R_2\) and \(R_1+R_3\to R_1\) yields
\[\left(\begin{array}{ccc|c}1&0&0&1-2j\\
0&1&0&2\\
0&0&1&-2j-1\end{array}\right)\]

Answer 6

Oops, the second matrix should be
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&j^2+j&0&2j+2j^2\\
0&j-1&-2&\color{red}{2j-4}\end{array}\right)\]
Then the next are
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&j-1&-2&\color{red}{2j-4}\end{array}\right)\]
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&j&-2&\color{red}{2j-2}\end{array}\right)\]
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&0&2&\color{red}2\end{array}\right)\]
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
0&1&0&2\\
0&0&1&\color{red}1\end{array}\right)\]
\[\left(\begin{array}{ccc|c}1&0&0&\color{red}{3}\\
0&1&0&2\\
0&0&1&\color{red}1\end{array}\right)\]
which agrees with your text.

Answer 7

Ok... can I do questions about?

Answer 8

Where you have
\[\left(\begin{array}{ccc|c}1&j&-1&2+2j\\
j&-j&-j&0\\
0&j-1&-2&-4+2j\end{array}\right)\]
You made a mistake in your second step, when you tried \(R_1+jR_2\to R_2\). Doing this would give
\[\left(\begin{array}{ccc|c}
1&j&-1&2+2j\\
\color{blue}{j^2+1}&\color{blue}{-j^2+j}&\color{blue}{-j^2-1}&\color{blue}{2+2j}\\
0&j-1&-2&-4+2j
\end{array}\right)\]

Answer 9

the first row have in the end 2+2j now I must subtract 6

thera are then 2+2j-6= -4+2j I cant understand yours it right I know but I cant understand.

Answer 10

I mean R1-R3

Answer 11

a ok...

Answer 12

I understand now. Ok but why the first line dont change when you multiplicate j must be the same?

Answer 13

The first row is not being replaced. If I have the row operation \(R_1+R_2\to R_2\), then only the second row, \(R_2\) is being replaced. No change is made to the first row.

Answer 14

a ok

Answer 15

Und how I do I it's completely false?

Answer 16

The row operations you used are not the most efficient, so the first thing I would do is work around those and find better/simpler ways to rewrite the given matrix. As a suggestion, try getting a matrix of the form
\[\left(\begin{array}{ccc|c}1&\cdots&\cdots&\cdots\\
0&\cdots&\cdots&\cdots\\
0&\cdots&\cdots&\cdots\end{array}\right)\]
then use the appropriate operation to get something that looks like
\[\left(\begin{array}{ccc|c}1&\cdots&\cdots&\cdots\\
0&1&\cdots&\cdots\\
0&0&\cdots&\cdots\end{array}\right)\]

Answer 17

\[\left(\begin{array}{ccc|c} 1&j&-1&2+2j\\ \color{blue}{j^2+1}&\color{blue}{-j^2+j}&\color{blue}{-j^2-1}&\color{blue}{2+2j}\\ 0&j-1&-2&-4+2j \end{array}\right)\]

Answer 18

this is my error but I dont understant because I cant do it?

Answer 19

hi

Answer 20

ok first question

Answer 21

\[\dfrac{1}{j^2+j}R_2\to R_2\]

Answer 22

waht mean this?

Answer 23

aslo 1 dived the j

Answer 24

Hi

Answer 25

hi

Answer 26

I checked your work, everything was fine until the last one.

Answer 27

You solved for z3=1, which is correct.

Answer 28

Sorry, I wasn't responding because I was check your matrices. It turns out that everything is fine, perhaps just a little problem with the back substitution.

To solve for z2, you can multiply both sides by 1-j that leaves
2z2=4, or z2=4
For z1, it would be a simple substitution.

If I didn't answer your question, send me a msg again in the morning. Must be quite late at your place. We just had supper.

Answer 29

*2z2=4, or z2=2

Answer 30

ok he explain me good bunt I onderstand some things from him not. when you can explain when you can.

Answer 31

\(\dfrac{1}{j^2+j}R_2\to R_2\)
is just another way of writing
\(\dfrac{1}{j-1}\)
or even
\(-\dfrac{1+j}{2}\)

Answer 32

ah ok you mean this (-1+j)z2=2j-z only multiply with 1-j ?

Answer 33

two not z

Answer 34

Yes, to solve for z2, you only have to multiply by the conjugate of the coefficient of z2, which is 1-j (or j-1 would do too).

Answer 35

The second line of the last matrix was
(1+j)z2 = 2+2j =2(1+j)
you can actually cancel (1+j) to get z2=2.

Answer 36

The general way to do it is to multiply by the conjugate to eliminate j.
(1-j)(1+j)z2=2(1+j)(1-j)
(1+1)z2=2(1+1)
so
z2=4/2=2

Answer 37

is -1+j ok but I understand complex coniugate

Answer 38

Yes, -1+j =-(1-j), so that would work too.

Answer 39

As long as it is a conjugate, and you multiply the same thing on both sides.

Answer 40

You're familiar with conjugates?

Answer 41

yes I must only the oposite of -1+j there are 1-j right?

Answer 42

and then multiply

Answer 43

exactly.

Answer 44

any complex number multiplied by the conjugate (Konjugat) will give a real number.
Example:
(4+3j)(4-3j)=16+9=25

Answer 45

Ok only one thing:

Answer 46

\[\dfrac{1}{j^2+j}R_2\to R_2\]

Answer 47

Yes?

Answer 48

you say there alternative wrien but in the matrix j^2+j became 1

Answer 49

He wanted to eliminate the coefficient (j^2+j) of z2, so he divided.

Answer 50

wrien = writing

Answer 51

He divided the line (j^2+j)z2 by (j^2+j) so it becomes 1z2.

Answer 52

Ok I understand thx for help with conjugate. I go sleep.

Answer 53

Very good! Talk another time. Gute nacht!

Answer 54

ok in my language. But my mother language are italy. can you say it?

Answer 55

is only joke.

Answer 56

Oh.. buona note?

Answer 57

yes note with two t notte

Answer 58

ok good night.

Answer 59

Oh yes, thank you and good night!

Answer 60

hi I must only see one thing no question.

Answer 61

\[\left(\begin{array}{ccc|c}1&j&2&-3+j\\ 3&j&j&-3\\ j&0&1&-1-j\end{array}\right)\]

Answer 62

z1+jz2+2z3=-3+j
3z1+jz2+jz3=-3
jz1 +z3=-1-j

Answer 63

R2-R1*j
Firs row dont change but I do auxiliary calculation:

3 3j 6 | -9+3j (this line is R1*j)
3 j j |-3

My problem is this part :

| -9+3j
|-3

I dont now when I must used bracket:

Answer 64

@franzmller682
If you are using gauß elimination, then you would be better off eliminating columns from left to right, i.e. do it in this sequence:
XYZ=P
XYZ=Q
XYZ=R
then
1YZ=P
0YZ=Q'
0YZ=R'
then
1YZ=P
01Z=Q''
00Z=R''
then
1YZ=P
01Z=Q''
001=R'''
Try not to worry about the vector on the right hand side.

I would start with 3R1-R2 -> R2 to get
R2: <0, 2j, 6-j | -6+3j>
and jR1-R3 -> R3 to get
R3: <0, -1, 2j-1 | -2j>

to get
\(\left(\begin{array}{ccc|c}1&j&2&-3+j\\ 0&2j&6-j&-6+3j\\ 0&-1&2j-1&-2j\end{array}\right)\)
and so on