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

  1. Jan 28, 2013 #1
    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 [itex] ad = bc [/itex], then the matrix [itex]\left [
    \begin{array}{cc}
    a & b \\
    c & d \\
    \end{array}
    \right][/itex] has no inverse.

    My attempt:

    Suppose there is an inverse such that

    [itex]\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]

    [/itex]

    By the definition of matrix multiplication, we have

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

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

    First let us suppose that [itex] a_{i} ≠ 0 [/itex] and seek contradiction.

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

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

    [itex] d = \frac{-ca_{i}}{c_{i}} [/itex]

    so

    [itex] ad = \frac{-ca_{i}}{c_{i}} * \frac{1-bc_{i}}{a_{i}} \\
    = \frac{-ca_{i}}{c_{i}} + bc
    [/itex]

    by substitution.

    Since we have by hypothesis [itex] ad = bc [/itex],

    [itex] bc = \frac{-ca_{i}}{c_{i}} + bc [/itex]

    so

    [itex] 0 = \frac{-ca_{i}}{c_{i}} [/itex].

    By supposition, [itex]a_{i}, c_{i} ≠ 0 [/itex], so, [itex] c = 0 [/itex]

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

    [itex]dd_{i} = 1[/itex], meaning that d ≠ 0,

    but we have from above that

    [itex] d = \frac{-ca_{i}}{c_{i}} = 0 [/itex].

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

    By (3),

    [itex] 0 = ca_{i} [/itex],

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

    Since [itex] ad = bc [/itex], either [itex]a[/itex] or [itex]d[/itex] must then be 0.

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

    [itex] 0 * a_{i} + 0 * b_{i} = 1 [/itex], a contradiction.

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

    [itex] 0 * b_{i} + 0 * d_{i} = 1 [/itex], a contradiction.

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

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

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

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

    Again by hypothesis,

    [itex] a * \frac{-ca_{i}}{c_{i}} = c * \frac{1-aa_{i}}{c_{i}} \\
    a * -ca_{i} = c * (1-aa_{i}) \\
    c = 0
    [/itex]

    We've already learned from above that c ≠ 0, so we're done.
     
    Last edited: Jan 28, 2013
  2. jcsd
  3. Jan 28, 2013 #2
    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.
     
  4. Jan 28, 2013 #3
    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
  5. Jan 28, 2013 #4

    Mark44

    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.
     
  6. Jan 28, 2013 #5
    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.
     
  7. Jan 28, 2013 #6
    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.
     
  8. Jan 28, 2013 #7
    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.
     
  9. Jan 28, 2013 #8
    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?
     
  10. Jan 28, 2013 #9

    rollingstein

    User Avatar
    Gold Member

    Is it just me that I find it odd that someone has heard of Bourbaki before the determinant? :surprised
     
  11. Jan 29, 2013 #10
    Thank you, LastOne. That was a great hint to a much better proof.

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

    [/itex]

    We can write that [itex] f(x,y) = (ax + by, cx + dy) [/itex].


    Finding [itex]f^{-1}[/itex] is the same thing as finding [itex]M^{-1}[/itex].

    Proof: [itex]
    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} \\
    [/itex]

    If we prove [itex]f^{-1}[/itex] doesn't exist, then we've proven that [itex]M^{-1}[/itex] 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.

    [itex] 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)
    [/itex]

    So, e.g., for

    M = [itex]\left [
    \begin{array}{cc}
    1 & 2 \\
    3 & 6 \\
    \end{array}
    \right] [/itex], 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
  12. Jan 29, 2013 #11
    "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
  13. Jan 29, 2013 #12
    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
  14. Jan 29, 2013 #13
    Doh. Fixed to a real example.
     
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