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

1. Jan 28, 2013

### 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.

Last edited: Jan 28, 2013
2. Jan 28, 2013

### 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.

3. Jan 28, 2013

### 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.

Last edited: Jan 28, 2013
4. Jan 28, 2013

### Staff: Mentor

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 ai, 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.

5. Jan 28, 2013

### 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.

6. Jan 28, 2013

### 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.

7. Jan 28, 2013

### 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.

8. Jan 28, 2013

### 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?

9. Jan 28, 2013

### rollingstein

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

10. Jan 29, 2013

### 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.

Last edited: Jan 29, 2013
11. Jan 29, 2013

### E'lir Kramer

"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.

Last edited: Jan 29, 2013
12. Jan 29, 2013

### VantagePoint72

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.

Last edited: Jan 29, 2013
13. Jan 29, 2013

### E'lir Kramer

Doh. Fixed to a real example.