If ad = bc, then the (abcd) matrix has no inverse

[E'lir Kramer]

This is problem 4.14 in Advanced Calculus of Several Variables. Two questions: Is my proof correct? And: is there a cleaner way to prove this?

Show that, if ad = bc, then the matrix \left [<br /> \begin{array}{cc}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{array}<br /> \right] has no inverse.

My attempt:

Suppose there is an inverse such that

\left [<br /> \begin{array}{cc}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{array}<br /> \right]<br /> <br /> \times<br /> <br /> \left [<br /> \begin{array}{cc}<br /> a_{i} &amp; b_{i} \\<br /> c_{i} &amp; d_{i} \\<br /> \end{array}<br /> \right]<br /> <br /> = I = <br /> <br /> \left [<br /> \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; 1 \\<br /> \end{array}<br /> \right]<br /> <br />

By the definition of matrix multiplication, we have

aa_{i} + bc_{i} = 1 \&gt;\&gt; (1) \\<br /> ab_{i} + bd_{i} = 0 \&gt;\&gt; (2) \\<br /> ca_{i} + dc_{i} = 0 \&gt;\&gt; (3) \\ <br /> cb_{i} + dd_{i} = 1 \&gt;\&gt; (4)<br />

Now, since aa_{i} + bc_{i} = 1, either a_{i} or c_{i} ≠ 0.

First let us suppose that a_{i} ≠ 0 and seek contradiction.

By our supposition and equations (1) and (3) above, we have
a = \frac{1-bc_{i}}{a_{i}} ≠ 0, also, c = \frac{-dc_{i}}{a_{i}}

To proceed further, we can also suppose that that c_{i} ≠ 0. Then we have

d = \frac{-ca_{i}}{c_{i}}

so

ad = \frac{-ca_{i}}{c_{i}} * \frac{1-bc_{i}}{a_{i}} \\<br /> = \frac{-ca_{i}}{c_{i}} + bc<br />

by substitution.

Since we have by hypothesis ad = bc,

bc = \frac{-ca_{i}}{c_{i}} + bc

so

0 = \frac{-ca_{i}}{c_{i}}.

By supposition, a_{i}, c_{i} ≠ 0, so, c = 0

But this is a contradiction, because then by (4),

dd_{i} = 1, meaning that d ≠ 0,

but we have from above that

d = \frac{-ca_{i}}{c_{i}} = 0.

Now if we suppose that a_{i} ≠ 0, c_{i} = 0, we are led to another contradiction.

By (3),

0 = ca_{i},

and by our first supposition that a_{i} ≠ 0, c = 0.

Since ad = bc, either a or d must then be 0.

If a = 0, we have by (1), and our supposition that c_{i} = 0, then

0 * a_{i} + 0 * b_{i} = 1, a contradiction.

If d = 0, we have by (4) and our supposition that c_{i} = 0, then

0 * b_{i} + 0 * d_{i} = 1, a contradiction.

Note that this proves that c ≠ 0 when c_{i} = 0. (We'll use this again in a second.)

Now we have proved that if a_{i} ≠ 0, we run into a contradiction.

The final case to consider is that a_{i} = 0. This is impossible. If a_{i} = 0, then c_{i} ≠ 0 by (1) . Then we have

b = \frac{1-aa_{i}}{c_{i}} ≠ 0 \&gt;\&gt; (1) \\<br /> d = \frac{-ca_{i}}{c_{i}} \&gt;\&gt; (3)<br />.

Again by hypothesis,

a * \frac{-ca_{i}}{c_{i}} = c * \frac{1-aa_{i}}{c_{i}} \\<br /> a * -ca_{i} = c * (1-aa_{i}) \\<br /> c = 0<br />

We've already learned from above that c ≠ 0, so we're done.

 

[Rikardus]

Matrix B has inverse iff det(B) is different than 0.
det B = ad -bc. ad = bc => det(B) = 0. Thus B has no inverse.

 

[E'lir Kramer]

I'll accept your proof, as long as you also prove that matrix B has an inverse iff det(B) ≠ 0 :)

Oh, and first, you'll have to define det() - we haven't covered it in the book. Edit: and by the way, I'm not intentionally playing dumb. I've never done linear algebra, and my first encounter with it is in this book. I really don't know what the det() function is.

 

[Mark44]

Since you haven't learned about determinants yet, then it is not reasonable to use them, as you point out.

Your proof seems very long-winded. You are doing a proof by contradiction in which you assume that ad = bc, and the the inverse of A exists. You should get a contradiction, which will mean that no such inverse exists.

You could simplify your work by using, say e, f, g, and h as the elements of the supposed inverse, instead of a_i, etc. When you row-reduce the matrix in the left half of your augmented matrix, you will at some point need to divide by ad - bc, which is zero by hypothesis.

 

[E'lir Kramer]

My tragic ignorance of linear algebra also leaves me bereft of any concept of "row reduction", as well :). I will say that the difficulty of this proof has motivated me to acquire the tools of linear algebra, though.

 

[VantagePoint72]

Linear algebra is pretty important for multivariable calculus. You really should get an introductory book for it. There doesn't seem much point in trying to work out what is purely a linear algebra problem without knowing linear algebra.

 

[E'lir Kramer]

Well, he's teaching me linear algebra right now, wouldn't you say? In the finest tradition of Bourbaki, he has yet to assign a problem that can't be solved without reference to external sources. After 60 harrowing, yet satisfying problems, I trust him to deliver me safely, as long as I do my part - and as long as I'm still getting traction on these end of chapter problems, I feel up to the challenge. Besides, I peeked ahead, and he's going to develop det() in ten pages.

 

[VantagePoint72]

Well, personally I think it would be ill-advised to learn linear algebra "on the fly" from a calculus book. However, for this particular problem, I suggest the easiest way to prove the statement without using any elementary linear algebra results is the following:

It is true in general (for linear and non-linear functions) that for a function to have an inverse, it must be one-to-one from its domain to its image. If your Socratically-minded textbook author hasn't established this yet, then it's a very convoluted book indeed. So, consider the function from $R^2$ to $R^2$ defined by left multiplying a column vector by the matrix you've given. (i.e. $f(x,y) = (ax + by, cx + dy)$). Argue that the matrix being invertible is equivalent to the function f having an inverse. Then construct, using the fact that $ad=bc$, two different column vectors that get mapped to the same vector by f. Of course, their existence means f is not 1-1 and thus not invertible.

There are infinitely many possible pairs of vectors you could use, though there are a few especially simple choices. Can you see what they are?

 

[rollingstein]

I'm not intentionally playing dumb. I've never done linear algebra, and my first encounter with it is in this book. I really don't know what the det() function is.

In the finest tradition of Bourbaki, he has yet to assign a problem that can't be solved without reference to external sources.

Is it just me that I find it odd that someone has heard of Bourbaki before the determinant?

 

[E'lir Kramer]

Thank you, LastOne. That was a great hint to a much better proof.

Considering the function f : \Re^{2} \to \Re^{2} such that f(x) = Mx where M = \left [<br /> \begin{array}{cc}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{array}<br /> \right]<br /> <br />

We can write that f(x,y) = (ax + by, cx + dy).Finding f^{-1} is the same thing as finding M^{-1}.

Proof: <br /> f^{-1}(f(x)) = (f^{-1} \circ f)(x) = M_{f^{-1}}M_{f}x \\<br /> <br /> (f^{-1} \circ f)(x) = x \\<br /> <br /> M_{f^{-1}}M_{f} x = x \\<br /> <br /> M_{f^{-1}}M_{f} = I \\<br /> <br /> M_{f^{-1}} = M_{f}^{-1} \\<br />

If we prove f^{-1} doesn't exist, then we've proven that M^{-1} doesn't. All that is required to prove that an inverse for f doesn't exist is to find two different values in the domain that are mapped to the same value by f.

If we also have that ad = bc, there are such vectors.

f((d-b),(a-c)) = (ad - ab + ab - bc, cd - bc + ad - cd) = (ad - bc, ad - bc) = (0, 0) \\<br /> f((b-d), (c-a)) = (ab - ad + bc - ab, bc - cd + cd - ad) = (bc - ad, bc - ad) = (0, 0)<br />

So, e.g., for

M = \left [<br /> \begin{array}{cc}<br /> 1 &amp; 2 \\<br /> 3 &amp; 6 \\<br /> \end{array}<br /> \right], we could think of f(4, -2) and f(-4, 2).

In general any vectors k(d - b, a - c) and k(b - d, c - a) map to zero.

 

[E'lir Kramer]

Is it just me that I find it odd that someone has heard of Bourbaki before the determinant?

"Bourbaki" was quoted at the beginning of a chapter in Spivak, and I looked "him" up the other day. I've been planning my own course of self study, so that entails some amount of scholarship of the history of mathematics. I identify with the project that those men were working on: as I understand it, in the aftermath of WWI, their teachers were dead or fled, and they were essentially resolved to teach themselves mathematics with books and elbow grease.

 

[VantagePoint72]

Thank you, LastOne. That was a great hint to a much better proof.

Indeed. You're welcome. I had in mind the slightly simpler case of f(d,0) = f(0,c) but of course yours works perfectly well too.

Edit: Though your specific example is wrong. For starters, $4 \not= 6$. I think the general case is probably fine, though I haven't worked it through.

 

[E'lir Kramer]

Doh. Fixed to a real example.