- #1

Tsunami317

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## Homework Statement

Determine the standard matrix for the linear operator defined by the formula below:

T(x, y, z) = (x-y, y+2z, 2x+y+z)

## Homework Equations

## The Attempt at a Solution

No idea

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- Thread starter Tsunami317
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- #1

Tsunami317

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Determine the standard matrix for the linear operator defined by the formula below:

T(x, y, z) = (x-y, y+2z, 2x+y+z)

No idea

- #2

HallsofIvy

Science Advisor

Homework Helper

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Hopefully, at least you know that, because this T is applied to a 3 component vector and the result is a 3 component vector, T will be represented by a 3 by 3 matrix.

And, you should know that applying T to a vector <x, y, z> is the same as the matrix multiplication

[tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}x \\ y \\z \end{bmatrix}[/tex]

Because they say the "standard" matrix, they want you to use the "standard" basis, <1, 0, 0>, <0, 1, 0>, and <0, 0, 1>. What do you get when you multiply the matrix above by each of those?

You are told that T(x, y, z) = (x-y, y+2z, 2x+y+z). What is T(1, 0, 0)? What is T(0, 1, 0)? What is T(0, 0, 1)?

Compare those with the result of multiplying the matrix.

- #3

Tsunami317

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- 0

[x-y 0 0

0 y+2z 0

0 0 2x+y+z]

?

or is the answer

[x-y

y+2x

2x+y+z]

I understand how to multiply matrices, but now this looks backward to me. Yes I am taking a distance learning class and there is no professor, just me reading a textbook which has terminology that rarely matches up with the sparse classnotes, let alone what I can google, and watching hundreds of hours of Khan Academy.

- #4

Tsunami317

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And thank you for your help!

- #5

- 9,568

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Because they say the "standard" matrix, they want you to use the "standard" basis, <1, 0, 0>, <0, 1, 0>, and <0, 0, 1>. What do you get when you multiply the matrix above by each of those?

You are told that T(x, y, z) = (x-y, y+2z, 2x+y+z). What is T(1, 0, 0)? What is T(0, 1, 0)? What is T(0, 0, 1)?

Compare those with the result of multiplying the matrix.

So the answer would be

[x-y 0 0

0 y+2z 0

0 0 2x+y+z]

?

or is the answer

[x-y

y+2x

2x+y+z]

No. It might help if you actually answer the questions Halls asked you above. Those answers might lead you to the solution.

- #6

Tsunami317

- 4

- 0

1 0 0

0 1 0

0 0 1

by the vectors given but I got this

[x-y

y+2x

2x+y+z]

Doesn't make sense

- #7

estro

- 241

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HallsofIvy and LCKurtz already provided great hints, I will repeat what they told you in other words:

You need to find a matrix A so that: A(x,y,z)=(x-y, y+2z, 2x+y+z)

It is important to understand the relationship between [linear] transformation and its "representative matrix", to be able to solve this problem.

With all the respect to Khan Academy I don't think it is the right place to learn linear algebra.

Try looking at: http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/

Prof. Gilbert Stran provides great lectures and excellent book.

You need to find a matrix A so that: A(x,y,z)=(x-y, y+2z, 2x+y+z)

It is important to understand the relationship between [linear] transformation and its "representative matrix", to be able to solve this problem.

With all the respect to Khan Academy I don't think it is the right place to learn linear algebra.

Try looking at: http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/

Prof. Gilbert Stran provides great lectures and excellent book.

Last edited:

- #8

- 9,568

- 774

1 0 0

0 1 0

0 0 1

by the vectors given but I got this

[x-y

y+2x

2x+y+z]

Doesn't make sense

Yes, it doesn't make sense, and it isn't what you were asked to do.

Halls asked you what you get when you multiply the matrix$$

\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$$and the three standard basis vectors which you would write as the matrices$$

\begin{bmatrix}1 \\0 \\ 0\end{bmatrix},\, \begin{bmatrix}0 \\ 1 \\0 \end{bmatrix}

,\, \begin{bmatrix}0 \\ 0 \\1 \end{bmatrix}$$What do you get when you multiply them? That is three different questions with three different answers, and there won't be any ##x,y,z## variables in the answer. Can you do that?

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