# matrices

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#1
can some please explain part a, and b? thanks
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1 year ago
#2
(Original post by FurtherMaths2020)
can some please explain part a, and b? thanks
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?
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1 year ago
#3
(Original post by FurtherMaths2020)
can some please explain part a, and b? thanks
a) Take any general point , apply the map to it, and argue that it lies on the line .

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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#4
(Original post by 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?
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.
Last edited by FurtherMaths2020; 1 year ago
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#5
(Original post by 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 ?
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.
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1 year ago
#6
(Original post by 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.
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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#7
(Original post by 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.
any linear transformation always map the origin onto itself, and any linear transformation can be represent by a matrix.
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1 year ago
#8
(Original post by FurtherMaths2020)
any linear transformation always map the origin onto itself, and any linear transformation can be represent by a matrix.
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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1 year ago
#9
(Original post by 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.
Is your matrix the right way round for b)?
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#10
(Original post by mqb2766)
Is your matrix the right way round for b)?
here, y = 6, and x = mb?

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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#11
(Original post by 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 .
what does it mean - a linear matrix always maps origin onto itself?
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1 year ago
#12
(Original post by FurtherMaths2020)
what does it mean - a linear matrix always maps origin onto itself?
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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#13
(Original post by mqb2766)
A matrix M always satisfies
0 = M0
So it maps the origin to the origin. The matrix just scales (and adds) individual entries.
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?
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#14
(Original post by 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, y = mx is a linear transformation and when x = 0, y = 0. when y = 0, x = 0? therefore, it passes through origin?
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1 year ago
#15
(Original post by 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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#16
(Original post by RDKGames)

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.
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)?
Last edited by FurtherMaths2020; 1 year ago
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1 year ago
#17
(Original post by 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)?
Most of what you've just said isn't very coherent, so sorry I don't even follow your argument.
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#18
(Original post by RDKGames)
Most of what you've just said isn't very coherent, so sorry I don't even follow your argument.
yeah, sorry - I meant: we know when a linear transformation is in the form of y = x, that when y = 0, x = 0, therefore it passes through the origin.

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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