Mixing
If $(a,b)$ is a multiple of $(c,d)$, show that $(a,c)$ is a multiple of $(b,d)$
$begingroup$
I need help with this problem:
If $(a,b)$ is a multiple of $(c,d)$ with $abcdneq0$, show that $(a,c)$ is a multiple of $(b,d)$. This is suprisingly important: call it a challenge question. You could use numbers first to see how $a,b,c$ and $d$ are related. The question will lead to:
If A = $left[begin{array}{l}a&b\c&dend{array}right]$ has dependent rows the it has dependent columns.
I tried to do it this way:
$(a,b)=x(c,d)Rightarrow(a,b)=(xc,xd)$
$(a,c)=(xc,c)Rightarrow c(x,1)$ and $(b,d)=(xd,d)Rightarrow d(x,1)$
I don't know what to do after that, what should I do next?
linear-algebra matrices
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
If $(a,b)$ is a multiple of $(c,d)$ with $abcdneq0$, show that $(a,c)$ is a multiple of $(b,d)$. This is suprisingly important: call it a challenge question. You could use numbers first to see how $a,b,c$ and $d$ are related. The question will lead to:
If A = $left[begin{array}{l}a&b\c&dend{array}right]$ has dependent rows the it has dependent columns.
I tried to do it this way:
$(a,b)=x(c,d)Rightarrow(a,b)=(xc,xd)$
$(a,c)=(xc,c)Rightarrow c(x,1)$ and $(b,d)=(xd,d)Rightarrow d(x,1)$
I don't know what to do after that, what should I do next?
linear-algebra matrices
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
$endgroup$
$begingroup$
Do you mean that $a=lambda c, b=lambda d$ for some constant $lambda$? But then $(a,c)=(lambda c, c)=ctimes (lambda ,1) $ and $(b,d)=(lambda d,d)=dtimes (lambda ,1)$ so...
$endgroup$
– lulu
Jan 27 at 17:30
$begingroup$
I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c over d}(b,d)$, am I right?
$endgroup$
– davidllerenav
Jan 27 at 19:15
$begingroup$
yes, that's right.
$endgroup$
– lulu
Jan 27 at 19:17
add a comment |
$begingroup$
I need help with this problem:
If $(a,b)$ is a multiple of $(c,d)$ with $abcdneq0$, show that $(a,c)$ is a multiple of $(b,d)$. This is suprisingly important: call it a challenge question. You could use numbers first to see how $a,b,c$ and $d$ are related. The question will lead to:
If A = $left[begin{array}{l}a&b\c&dend{array}right]$ has dependent rows the it has dependent columns.
I tried to do it this way:
$(a,b)=x(c,d)Rightarrow(a,b)=(xc,xd)$
$(a,c)=(xc,c)Rightarrow c(x,1)$ and $(b,d)=(xd,d)Rightarrow d(x,1)$
I don't know what to do after that, what should I do next?
linear-algebra matrices
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
$endgroup$
I need help with this problem:
If $(a,b)$ is a multiple of $(c,d)$ with $abcdneq0$, show that $(a,c)$ is a multiple of $(b,d)$. This is suprisingly important: call it a challenge question. You could use numbers first to see how $a,b,c$ and $d$ are related. The question will lead to:
If A = $left[begin{array}{l}a&b\c&dend{array}right]$ has dependent rows the it has dependent columns.
I tried to do it this way:
$(a,b)=x(c,d)Rightarrow(a,b)=(xc,xd)$
$(a,c)=(xc,c)Rightarrow c(x,1)$ and $(b,d)=(xd,d)Rightarrow d(x,1)$
I don't know what to do after that, what should I do next?
linear-algebra matrices
linear-algebra matrices
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
asked Jan 27 at 17:27
davidllerenavdavidllerenav
3078
3078
$begingroup$
Do you mean that $a=lambda c, b=lambda d$ for some constant $lambda$? But then $(a,c)=(lambda c, c)=ctimes (lambda ,1) $ and $(b,d)=(lambda d,d)=dtimes (lambda ,1)$ so...
$endgroup$
– lulu
Jan 27 at 17:30
$begingroup$
I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c over d}(b,d)$, am I right?
$endgroup$
– davidllerenav
Jan 27 at 19:15
$begingroup$
yes, that's right.
$endgroup$
– lulu
Jan 27 at 19:17
add a comment |
$begingroup$
Do you mean that $a=lambda c, b=lambda d$ for some constant $lambda$? But then $(a,c)=(lambda c, c)=ctimes (lambda ,1) $ and $(b,d)=(lambda d,d)=dtimes (lambda ,1)$ so...
$endgroup$
– lulu
Jan 27 at 17:30
$begingroup$
I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c over d}(b,d)$, am I right?
$endgroup$
– davidllerenav
Jan 27 at 19:15
$begingroup$
yes, that's right.
$endgroup$
– lulu
Jan 27 at 19:17
$begingroup$
Do you mean that $a=lambda c, b=lambda d$ for some constant $lambda$? But then $(a,c)=(lambda c, c)=ctimes (lambda ,1) $ and $(b,d)=(lambda d,d)=dtimes (lambda ,1)$ so...
$endgroup$
– lulu
Jan 27 at 17:30
$begingroup$
Do you mean that $a=lambda c, b=lambda d$ for some constant $lambda$? But then $(a,c)=(lambda c, c)=ctimes (lambda ,1) $ and $(b,d)=(lambda d,d)=dtimes (lambda ,1)$ so...
$endgroup$
– lulu
Jan 27 at 17:30
$begingroup$
I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c over d}(b,d)$, am I right?
$endgroup$
– davidllerenav
Jan 27 at 19:15
$begingroup$
I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c over d}(b,d)$, am I right?
$endgroup$
– davidllerenav
Jan 27 at 19:15
$begingroup$
yes, that's right.
$endgroup$
– lulu
Jan 27 at 19:17
$begingroup$
yes, that's right.
$endgroup$
– lulu
Jan 27 at 19:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: If $ad=bc$ then $dfrac ab=dfrac cd$ and also $dfrac ac=dfrac bd$
answered Jan 27 at 17:33
zwimzwim
12.6k831
$endgroup$
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
1
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
add a comment |
$begingroup$
First note the following: $(a,b) = (a,(frac{b}{a}) a)$. [As $abcd not =0$ we can assume that $frac{b}{a}$ exists]
So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(frac{b}{a}) a$.
Thus $(b,d) = (frac{b}{a} a, x(frac{b}{a})a) = frac{b}{a}(a,xa) = frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = frac{b}{a}$.
answered Jan 27 at 17:33
MikeMike
4,461512
$endgroup$
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
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$begingroup$
Hint: If $ad=bc$ then $dfrac ab=dfrac cd$ and also $dfrac ac=dfrac bd$
answered Jan 27 at 17:33
zwimzwim
12.6k831
$endgroup$
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
1
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
add a comment |
$begingroup$
Hint: If $ad=bc$ then $dfrac ab=dfrac cd$ and also $dfrac ac=dfrac bd$
answered Jan 27 at 17:33
zwimzwim
12.6k831
$endgroup$
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
1
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
add a comment |
$begingroup$
Hint: If $ad=bc$ then $dfrac ab=dfrac cd$ and also $dfrac ac=dfrac bd$
answered Jan 27 at 17:33
zwimzwim
12.6k831
$endgroup$
Hint: If $ad=bc$ then $dfrac ab=dfrac cd$ and also $dfrac ac=dfrac bd$
answered Jan 27 at 17:33
zwimzwim
12.6k831
answered Jan 27 at 17:33
zwimzwim
12.6k831
answered Jan 27 at 17:33
zwimzwim
12.6k831
answered Jan 27 at 17:33
zwimzwim
12.6k831
12.6k831
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
1
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
add a comment |
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
1
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $frac ac=frac bd$ is the hypothesis, and $frac ab=frac cd$ is the conclusion
$endgroup$
– Arthur
Jan 27 at 17:36
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0.
$endgroup$
– zwim
Jan 27 at 17:41
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
$begingroup$
You have to be careful lest you end up dividing by 0.
$endgroup$
– Mike
Jan 27 at 17:42
1
1
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
@Mike OP specified $abcdneq 0$.
$endgroup$
– zwim
Jan 27 at 17:42
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
$begingroup$
You are right, $abcd not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either
$endgroup$
– Mike
Jan 27 at 17:44
add a comment |
$begingroup$
First note the following: $(a,b) = (a,(frac{b}{a}) a)$. [As $abcd not =0$ we can assume that $frac{b}{a}$ exists]
So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(frac{b}{a}) a$.
Thus $(b,d) = (frac{b}{a} a, x(frac{b}{a})a) = frac{b}{a}(a,xa) = frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = frac{b}{a}$.
answered Jan 27 at 17:33
MikeMike
4,461512
$endgroup$
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
add a comment |
$begingroup$
First note the following: $(a,b) = (a,(frac{b}{a}) a)$. [As $abcd not =0$ we can assume that $frac{b}{a}$ exists]
So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(frac{b}{a}) a$.
Thus $(b,d) = (frac{b}{a} a, x(frac{b}{a})a) = frac{b}{a}(a,xa) = frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = frac{b}{a}$.
answered Jan 27 at 17:33
MikeMike
4,461512
$endgroup$
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
add a comment |
$begingroup$
First note the following: $(a,b) = (a,(frac{b}{a}) a)$. [As $abcd not =0$ we can assume that $frac{b}{a}$ exists]
So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(frac{b}{a}) a$.
Thus $(b,d) = (frac{b}{a} a, x(frac{b}{a})a) = frac{b}{a}(a,xa) = frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = frac{b}{a}$.
answered Jan 27 at 17:33
MikeMike
4,461512
$endgroup$
First note the following: $(a,b) = (a,(frac{b}{a}) a)$. [As $abcd not =0$ we can assume that $frac{b}{a}$ exists]
So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(frac{b}{a}) a$.
Thus $(b,d) = (frac{b}{a} a, x(frac{b}{a})a) = frac{b}{a}(a,xa) = frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = frac{b}{a}$.
answered Jan 27 at 17:33
MikeMike
4,461512
edited Jan 27 at 17:46
answered Jan 27 at 17:33
MikeMike
4,461512
answered Jan 27 at 17:33
MikeMike
4,461512
answered Jan 27 at 17:33
MikeMike
4,461512
4,461512
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
add a comment |
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
So I just need to make that algebraic trick, and that's all, right?
$endgroup$
– davidllerenav
Jan 27 at 19:08
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
$begingroup$
Yes that is correct.
$endgroup$
– Mike
Jan 28 at 1:29
add a comment |
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$begingroup$
Do you mean that $a=lambda c, b=lambda d$ for some constant $lambda$? But then $(a,c)=(lambda c, c)=ctimes (lambda ,1) $ and $(b,d)=(lambda d,d)=dtimes (lambda ,1)$ so...
$endgroup$
– lulu
Jan 27 at 17:30
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I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c over d}(b,d)$, am I right?
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– davidllerenav
Jan 27 at 19:15
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yes, that's right.
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– lulu
Jan 27 at 19:17