# matrices

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#2

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can some please explain part a, and b? thanks

**FurtherMaths2020**)can some please explain part a, and b? thanks

b) The new matrix must do something similar to a) bit introduces a multiplier "m". Any ideas?

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#3

(Original post by

can some please explain part a, and b? thanks

**FurtherMaths2020**)can some please explain part a, and b? thanks

b) Use the matrix in part (a) but adjust it slightly so that it maps the general point onto .

c) A linear map must satisfy two conditions. What are they? Can you show that one of them fails if any point goes to ?

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(Original post by

a) Consider the matrix multiplying the column vector (x1,y1). What is the result?

b) The new matrix must do something similar to a) bit introduces a multiplier "m". Any ideas?

**mqb2766**)a) Consider the matrix multiplying the column vector (x1,y1). What is the result?

b) The new matrix must do something similar to a) bit introduces a multiplier "m". Any ideas?

as for b, replace (1,1) with (m ,1), and I get ( bm, b)?

whats going on with c? I dont understand what the question is saying, nor asking.

Last edited by FurtherMaths2020; 1 year ago

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(Original post by

c) A linear map must satisfy two conditions. What are they?

**RDKGames**)c) A linear map must satisfy two conditions. What are they?

**Can you show that one of them fails if any point goes to ?**for c: for the line to be linear, the gradient must be constant. no, I don't know how to show that.

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#6

(Original post by

thanks for part a and b,

for c: for the line to be linear, the gradient must be constant. no, I don't know how to show that.

**FurtherMaths2020**)thanks for part a and b,

for c: for the line to be linear, the gradient must be constant. no, I don't know how to show that.

Go back through your textbook. There *must* be a definition for a linear transformation / mapping in there that you should know.

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(Original post by

No that's not it.

Go back through your textbook. There *must* be a definition for a linear transformation / mapping in there that you should know.

**RDKGames**)No that's not it.

Go back through your textbook. There *must* be a definition for a linear transformation / mapping in there that you should know.

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#8

(Original post by

any linear transformation always map the origin onto itself, and any linear transformation can be represent by a matrix.

**FurtherMaths2020**)any linear transformation always map the origin onto itself, and any linear transformation can be represent by a matrix.

and

But fine, you can use the fact that the origin must map back onto itself.

So go ahead and apply the map to the origin and argue that the result is not on the line where .

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#9

(Original post by

ok, for part a, if I multiply a by (a, b) I get (b, b), in which case, y = x.

as for b, replace (1,1) with (m ,1), and I get ( bm, b)?

whats going on with c? I dont understand what the question is saying, nor asking.

**FurtherMaths2020**)ok, for part a, if I multiply a by (a, b) I get (b, b), in which case, y = x.

as for b, replace (1,1) with (m ,1), and I get ( bm, b)?

whats going on with c? I dont understand what the question is saying, nor asking.

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(Original post by

Is your matrix the right way round for b)?

**mqb2766**)Is your matrix the right way round for b)?

ok, I see where I've gone wrong. Y = mb, and x = b, in the equation: y = mx

so, ive put the m in the wrong place.

Last edited by FurtherMaths2020; 1 year ago

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(Original post by

I was thinking more along the lines of two conditions ... T is linear if:

and

But fine, you can use the fact that the origin must map back onto itself.

So go ahead and apply the map to the origin and argue that the result is not on the line where .

**RDKGames**)I was thinking more along the lines of two conditions ... T is linear if:

and

But fine, you can use the fact that the origin must map back onto itself.

So go ahead and apply the map to the origin and argue that the result is not on the line where .

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#12

(Original post by

what does it mean - a linear matrix always maps origin onto itself?

**FurtherMaths2020**)what does it mean - a linear matrix always maps origin onto itself?

0 = M0

So it maps the origin to the origin. The matrix just scales (and adds) individual entries.

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(Original post by

A matrix M always satisfies

0 = M0

So it maps the origin to the origin. The matrix just scales (and adds) individual entries.

**mqb2766**)A matrix M always satisfies

0 = M0

So it maps the origin to the origin. The matrix just scales (and adds) individual entries.

so, c must equal c? because otherwise, the transformation that satisfies that equation doesnt pass through the origin and doesnt scale individual entries, but moves them?

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**mqb2766**)

A matrix M always satisfies

0 = M0

So it maps the origin to the origin. The matrix just scales (and adds) individual entries.

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#15

(Original post by

ah, ok!

so, c must equal c? because otherwise, the transformation that satisfies that equation doesnt pass through the origin and doesnt scale individual entries, but moves them?

**FurtherMaths2020**)ah, ok!

so, c must equal c? because otherwise, the transformation that satisfies that equation doesnt pass through the origin and doesnt scale individual entries, but moves them?

A linear map must map (0,0) to (0,0). So if, after the transformation, you have (0,0) you should check whether it is on the line . Is it?

Well, subbing the coordinates in you get .

So for a general this condition does not hold. Therefore the transformation is generally not linear.

Last edited by RDKGames; 1 year ago

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(Original post by

It's not too difficult, let's be clear about this ...

A linear map must map (0,0) to (0,0). So if, after the transformation, you have (0,0) you should check whether it is on the line . Is it?

Well, subbing the coordinates in you get .

So for a general this condition does not hold. Therefore the transformation is generally not linear.

**RDKGames**)It's not too difficult, let's be clear about this ...

A linear map must map (0,0) to (0,0). So if, after the transformation, you have (0,0) you should check whether it is on the line . Is it?

Well, subbing the coordinates in you get .

So for a general this condition does not hold. Therefore the transformation is generally not linear.

if it doesnt pass through the origin, then: ax + by = c, when x = 0, y = c, not zero (unless c is 0), therefore, ax + by = c is not a linear transformation unless c = 0 (passes through origin)?

Last edited by FurtherMaths2020; 1 year ago

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#17

(Original post by

ok, it must be linear and pass through the origin. so when y = x, and when x =0, y =0.

if it doesnt pass through the origin, then: ax + by = c, when x = 0, y = c, not zero (unless c is 0), therefore, ax + by = c is not a linear transformation unless c = 0 (passes through origin)?

**FurtherMaths2020**)ok, it must be linear and pass through the origin. so when y = x, and when x =0, y =0.

if it doesnt pass through the origin, then: ax + by = c, when x = 0, y = c, not zero (unless c is 0), therefore, ax + by = c is not a linear transformation unless c = 0 (passes through origin)?

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(Original post by

Most of what you've just said isn't very coherent, so sorry I don't even follow your argument.

**RDKGames**)Most of what you've just said isn't very coherent, so sorry I don't even follow your argument.

in the form of y = x + c, c must equal 0 (pass through origin), otherwise, when x = 0, y doesn't equal 0, but c.

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