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Question: Show that if $A$ is invertible and $AB = AC$, then $B = C$.

My work: My thought process: If I can find the inverse of $A$, then I can show A is invertible.

I will prove by example. $A$ is a $2 \times 2$ matrix. first row: 2 3. second row: 4 5. Where $a = 2$, $b = 3$, $c = 4$, $d = 5$ such that $ad - bc$ is not zero. $ad-bc$ or the determinant is $-2$ which is not equal to zero so I know matrix $A$ is invertible.

$A$ inverse would be...first row is -5/2 3/2. second row is 2 -1.

Matrix $C$ is equal to $A$ inverse or matrix $B$. $A$ inverse and matrix $B$ are the same thing.

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    $\begingroup$ Note that $A^{-1}A=I$... $\endgroup$ – Nick Mar 3 '14 at 0:42
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    $\begingroup$ You're not asked to show that $A$ is invertible, you are given that as an assumption. $\endgroup$ – Robert Israel Mar 3 '14 at 0:42
  • $\begingroup$ @Nick I did multiply the matrix A by its inverse and got the identity matrix. $\endgroup$ – Nicholas Mar 3 '14 at 0:45
  • $\begingroup$ What is $A^{-1}AB$? $\endgroup$ – Nick Mar 3 '14 at 0:45
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    $\begingroup$ By definition, if $A$ is invertible then there exists a matrix, called $A^{-1}$, such that $AA^{-1}=A^{-1}A=I$, where $I$ is the identity matrix. And $IB=B$ for any matrix $B$ such that the product is defined. So $A^{-1}AB=IB=B$. $\endgroup$ – Nick Mar 3 '14 at 0:52
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Generally speaking, you can't prove anything by example. You can prove all integers are even if you are allowed to "prove by example". I pick 2, 98, and 600! See? Everything I tried, worked! But that doesn't tell you anything about "all integers". So, it's not enough to show that it works in one or two or a hundred cases.

[On the other hand, working out a few examples can be extremely beneficial in an effort to figure out why some idea X has to be true in all cases, and can lead you to a proof.]

In this case, however, you're looking at an example of why they say that "math is 40% writing 1 in a creative way, and 40% writing zero in a creative way".

The trick with this one is what you might call a "hidden given". They gave us that A is invertible, and they gave us that AB = AC. But what is the hidden given? The hidden given is that B is equal to itself. Very tricky, and not immediately obvious how it helps us.

But this is what we're going to do with it: we know that B = B. We're going to start with that equality, and play around mathematically until it looks like what we want to prove (namely, that B = C).

So, here we go:

B = B

How can we get an A into that equation? There's a reason for us to want to do that. AB is equal to AC. If we can put an A in that equation somehow, we might then be able to get a C in there, and then maybe we can get it to tell us that B = C.

Well, here's where we use the rule that much of math is "creatively writing one". One times B is B--in matrix-ese, that is "IB=B". Your brain has to make the leap that--since we can also write I as A-1A--this is our ticket to getting the A in there where once we only had the boring fact that B=B.

To be formal, we "know that IB = B, by definition of the identity". And when they say that A is invertible, in formal terms that means that we know that "there exists an A-1 such that A-1A = I and AA-1 = I".

So, in this little equation we get from applying the definition of the identity to "B = B":

IB = B

, we're going to creatively write I as A-1A.

(A-1A)B = B

Note that all we've done here is to use the definition of the identity and the definition of an inverse--because of what the identity means, we can write IB=B. And because of what invertible means, we can write A-1A = I.

Now we're going to use associativity--the fact that matrix multiplication can be grouped however we want. The fact that for any X, Y, and Z, (XY)Z is the same as X(YZ). So, going back to where we were

(A-1A)B = B

we can regroup that stuff on the left like this:

A-1(AB) = B

Why would we do that? Because it gives us an (AB) to work with. And they told us something about AB! What did they tell us? That AB=AC. Well, if (AB) is equal to (AC), when we can put that AC right in there for (AB).

A-1(AC) = B

So, to recap where we are--they told us AB=AC. And they told us that A is invertible. And we took those facts, and used them together with a bunch of properties and definitions: (1) what invertible means (2) definition of the identity (3) matrix multiplication is associative (we an even add (0) that every matrix is equal to itself--"equality is reflexive", is how you say that formally).

Well, looking at A-1(AC) = B, it's just screaming at us to apply associativity again. Let's group up that A-1 with its old buddy A again, shall we?

(A-1A)C = B

Now, then, the definition of an inverse tells us that A-1A is I. Let's apply that:

IC = B

Now let's apply the definition of identity. That "IC" we've got is just equal to C, so:

C = B

It's beneficial to try to get an idea of where you needed the givens. If A wasn't invertible, what would go wrong?

Well, suppose A was the zero matrix (which is not invertible). If A is the zero matrix, then knowing that AB = AC doesn't necessarily tell you anything about B and C--you could literally put any B and C in there, and the equality would still hold. So you need the fact that A is invertible if you want to go from AB = AC to B = C. Your high school algebra brain just screams at you "Cancel the A's! CANCEL THE A's NOW!". But it was never ok to divide both sides by anything unless you were sure that the "anything" wasn't zero. And in matrix-land, things are a little more complex, but there's a similar rule. If it's not invertible, you don't know if you can "cancel" it or not. And that's really what this exercise is about--it's about learning when you can go from AB = AC to B = C. And the answer to that is, "when A is invertible".

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    $\begingroup$ Take out the editorializing and your Answer will be much improved. $\endgroup$ – hardmath Mar 3 '14 at 3:01
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    $\begingroup$ Toooooo long, tooooo wordy and, eventually, too boring. When things are simple do not, please, do not make them twisted: $$AB=AC\implies A^{-1}(AB)=A^{-1}(AC)\stackrel{\text{assoc.}}\implies$$ $$\left(A^{-1}A\right)B=\left(A^{-1}A\right)C\implies IB=IC\implies B=C$$ Anyway, I'm upvoting this answer for the effort and because it is a newie. +1 $\endgroup$ – DonAntonio Mar 3 '14 at 3:47
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    $\begingroup$ I appreciate your comment. The length of the post is due to the fact that I didn't want to merely do the asker's homework for him. Since I was going to effectively do the homework, I felt that the asker would benefit from knowing how one might have thought of this particular approach. Hopefully asker will read through and gain some insight on the nature of the problem, and why the problem was posed to them, rather than merely saying "Oh! Why didn't I think of that?" and writing it down and handing it in. $\endgroup$ – msouth Mar 4 '14 at 21:03

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