Question

linear transformations.

Answer 1

What is your question?

Answer 2

http://www.ping.be/~ping1339/lintf.htm

Answer 3

Let \[T:R^{4} \rightarrow R^{3}\]
T(x,y,z,t)=( x+3y+5z-2t, 3x-2y-7z+5t, 2x+y+t )
Solve the linear problem
\[ T( \xi )=\beta\] where
\[\beta= (11,0,7)\]

Answer 4

@beketso is the best link that can clear all yur doubts that you have related linear transformation...

Answer 5

It is a question to solve three equations in 4 unknown

Answer 6

Try to find x ,y and z in terms of t

Answer 7

do you mean
x+3y+5z=2t
3x-2y-7z=-5t
2x+y=-t

Answer 8

almost there. Do not forget \(\beta \)

Answer 9

i get x=t ,y=-2t ,z=t.what do i do with beta?

Answer 10

x+3y+5z=2t +11
3x-2y-7z=-5t+0
2x+y=7-t

Answer 11

i now get x=t+2 ,y=-2t+3 , z=t

Answer 12

@eliassaab ,why did you add coordinates of beta to each equation?

Answer 13

I am now getting confused .i'm getting
x=2 ,y=3-t , z=t

Answer 14

OK I see that people have already guided you but abandoned you. Sorry about that. It's sad that people can't finish the help they started to give, leaving the person needing help stranded. They haven't misguided you however. Thus we will continue where they left off. Let's suppose we begin with you giving me an idea of where you may be lost.

Answer 15

is my last post right?

Answer 16

Give me a minute. I will check it.

Answer 17

Your x, y, and z in terms of t are not correct. If you show me your steps which led you to this conclusion, then I can show the careless errors which may have resulted in confusion.

Answer 18

okay

Answer 19

1 3 5 -2 | 11
3 -2 -7 5 | 0
2 1 0 1 | 7

Answer 20

then i reduce the matrix

Answer 21

1 3 5 -2 | 11
0 -11 -22 11|-33
0 5 -10 5|-15

Answer 22

Since you desire to solve for x, y, and z in terms of t, the t then should be part of the constant matrix rather than the coefficient matrix.

Answer 23

what is the constant matrix?

Answer 24

The coefficient matrix which is to the left of the imaginary line | of the augmented matrix contains the numerical coefficients of the variables x, y, and z on the left side of the equations. The constant matrix is the matrix to the right of the imaginary line | of the augmented matrix and consists of the constants on the right side of the equations.

Answer 25

Here t is treated as a constant and thus is part of the constant matrix.

Answer 26

if t is part of the constant matrix,the do i add the coefficients of t to the coordinate of beta?

Answer 27

Yes so then to the right of | you have

11 + 2t
- 5t
7 - t

Answer 28

will i have
1 3 5 |13
3 -2 -7 |-5
2 1 0 |-6

Answer 29

No, you shall have

1 3 5 | 11 + 2t
3 -2 -7 | -5t
2 1 0 | 7 - t

Answer 30

then i solve for x,y and z and then get the answer in terms of t?

Answer 31

Yes but first it shall be easier if column 1 and column 3 are interchanged.

Answer 32

@calculusfunctions ,thank u a lot.i'm solving right now

Answer 33

No thanks needed! Your correct solution and understanding of the concepts so that you may succeed in solving future problems independently, are thanks enough.

Answer 34

my augmented matrix reduces to
1 0 1 |2-t
0 1 2 |3+t
0 0 0 |0

Answer 35

meaning x=2+z-t, y=3-2z+t

Answer 36

made a mistake my matrix reduces to
1 0 -1 |2-t
0 1 2 |t+3
0 0 0 |0

Answer 37

If the last row is a zero, then 0z = 0 so then z = k; k ∈ ℝ.

Answer 38

that will mean
x=2+k-t ,y=3-2k+t and z=k

Answer 39

I haven't actually checked if your final answer is correct. Give me a few minutes and I will do so.

Answer 40

@REMAINDER

Answer 41

@beketso sorry but I was double checking the tedious calculations and your answer is incorrect.

Answer 42

I was also solving it myself which is why it took some time.

The correct answer should be

x = k
y = 7 - 2k - t
z = -2 + k + t

Answer 43

okay.i will try to do it again.does the fact that row 3 was exchanged with row 1 affect the solution?

Answer 44

Yes it does.

Answer 45

@beketso do you need me to stick around or are you now feeling confident?

Answer 46

Since there is a non-response, I shall assume my help is no longer needed here. Bye!

Answer 47

@calculusfunctions thank you