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 $\left [ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right]$ has no inverse.

My attempt:

Suppose there is an inverse such that

$\left [ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] \times \left [ \begin{array}{cc} a_{i} & b_{i} \\ c_{i} & d_{i} \\ \end{array} \right] = I = \left [ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right]$

By the definition of matrix multiplication, we have

$aa_{i} + bc_{i} = 1 \>\> (1) \\ ab_{i} + bd_{i} = 0 \>\> (2) \\ ca_{i} + dc_{i} = 0 \>\> (3) \\ cb_{i} + dd_{i} = 1 \>\> (4)$

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

$ad = \frac{-ca_{i}}{c_{i}} * \frac{1-bc_{i}}{a_{i}} \\ = \frac{-ca_{i}}{c_{i}} + bc$

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

$b = \frac{1-aa_{i}}{c_{i}} ≠ 0 \>\> (1) \\ d = \frac{-ca_{i}}{c_{i}} \>\> (3)$.

Again by hypothesis,

$a * \frac{-ca_{i}}{c_{i}} = c * \frac{1-aa_{i}}{c_{i}} \\ a * -ca_{i} = c * (1-aa_{i}) \\ c = 0$

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 = $\left [ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right]$

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

Proof: $f^{-1}(f(x)) = (f^{-1} \circ f)(x) = M_{f^{-1}}M_{f}x \\ (f^{-1} \circ f)(x) = x \\ M_{f^{-1}}M_{f} x = x \\ M_{f^{-1}}M_{f} = I \\ M_{f^{-1}} = M_{f}^{-1} \\$

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.

$f((d-b),(a-c)) = (ad - ab + ab - bc, cd - bc + ad - cd) = (ad - bc, ad - bc) = (0, 0) \\ f((b-d), (c-a)) = (ab - ad + bc - ab, bc - cd + cd - ad) = (bc - ad, bc - ad) = (0, 0)$

So, e.g., for

M = $\left [ \begin{array}{cc} 1 & 2 \\ 3 & 6 \\ \end{array} \right]$, 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.