Does the Inverse Function Matrix Always Have a Non-Zero Determinant?

[yeongil]

A friend of mine is tutoring a student in math, and they came across a problem that she couldn't answer. Here's a problem: both she and the student are Korean, and her English is okay, but not so good when it comes to specific terminology (like math). My Korean is very limited. The problem, if I understand her correctly, is below. Can anyone check my work to see if this is correct?

Homework Statement

Given a function f such that f(a) = b and f^-1(c) = d [f inverse of c is d]. Construct a 2x2 matrix A like this:
a b
c d
For each of the following functions, does A^-1 always exist?

Homework Equations

(1) y = x + 1
(2) y = log(x)
(3) y = 2^x
(4) y = sqrt(x - 1)
(5) y = 1/x

The Attempt at a Solution

Here are the inverses of the functions above:
(1) y = x - 1
(2) y = 10^x
(3) y = log₂(x)
(4) y = x² + 1 (for x>=0 only)
(5) y = 1/x

A^-1 exists if the determinant is not zero, ie. ad - bc <> 0.
Let's look at cases where ad - bc = 0.
Given the definitions above, are the only ways that the determinant is zero is if
(A) the ordered pairs (a, b) and (c, d) are the same, or
(B) c = ka and d = kb, for some number k?

Case (A): (a, b) = (c, d)
If the ordered pairs are the same, that means f(x) and f^-1(x) must intersect at that point, right?
If f(x) and f^-1(x) are not the same function, and if they intersect at all, they can only do so at points on the y = x line, correct?

Case (B): (c, d) = (ka, kb)
If c = ka and d = kb, for some number k, is there anything I can generalize here?

For functions (1) - (4), none of them and their inverses intersect (case A), and I can't find an ordered pair in the original functions so that the multiple of their coordinates satisfy the inverses functions (case B). Is it safe to say that so for those, A^-1 always exist?

For function (5), f(x) = f^-1(x), so any ordered pair in f(x) will also hold true in f^-1(x). So for Case A, the determinant is 0, so A^-1 won't exist. I can find plenty of pairs of points that would satisfy case B [like (2, 1/2) and (4, 1/4)], making the determinant 0, which means that A^-1 won't exist.

So function (5) is the only one that doesn't work. Am I right? Is there anything that I'm missing? Has anyone seen a problem like this? TIA for checking.


01

 

[tiny-tim]

Welcome to PF!

Given a function f such that f(a) = b and f^-1(c) = d [f inverse of c is d]. Construct a 2x2 matrix A like this:
a b
c d
For each of the following functions, does A^-1 always exist?

Homework Equations

(1) y = x + 1
(2) y = log(x)
(3) y = 2^x
(4) y = sqrt(x - 1)
(5) y = 1/x

Hi yeongil! Welcome to PF!

I've only looked at (1) y = x + 1, and I've noticed that:

a a+1
-a -a-1

has determinant zero, and no inverse.

Does that help you with the others?

 

[yeongil]

Hi yeongil! Welcome to PF!

I've only looked at (1) y = x + 1, and I've noticed that:

a a+1
-a -a-1

has determinant zero, and no inverse.

Does that help you with the others?

Thanks for the welcome.

After looking over (2) - (4), I've noticed a few more things.

ad - bc = 0 -> ad = bc -> a/b = c/d [provided that b <> 0 and d <> 0]

(2) f(a) = b = log(a)
f^-1(c) = d = 10^c
a/b = c/d
a/log(a) = c/10^c

I was able to find values for a and c that would make the proportion true, using a calculator and brute force. But is that the only way to do it? Anyway, since I found values that made the determinant of A zero, (2) doesn't work either.

(3) f(a) = b = 2^a
f^-1(c) = d = log₂(c)
a/b = c/d
a/2^a = c/log₂(c)

Did it the same way as (2). Found values using the calculator that made the determinant of A zero, so (3) doesn't work either.

(4) f(a) = b = sqrt(a - 1)
f^-1(c) = d = c² + 1 [c >= 0]
a/b = c/d
a/sqrt(a - 1) = c/(c² + 1) [c >= 0]
Not possible.
For all a > 1, a/b always >= 2
For all c >= 0, c/d always <= 0.5
So the determinant is never zero. (4) works.

Anything else I'm missing? TIA.


01

 

[QuantumKing]

Thanks for the welcome.

After looking over (2) - (4), I've noticed a few more things.

ad - bc = 0 -> ad = bc -> a/b = c/d [provided that b <> 0 and d <> 0]

(2) f(a) = b = log(a)
f^-1(c) = d = 10^c
a/b = c/d
a/log(a) = c/10^c

I was able to find values for a and c that would make the proportion true, using a calculator and brute force. But is that the only way to do it? Anyway, since I found values that made the determinant of A zero, (2) doesn't work either.


01

Well, if you rearrange (2) from a/log(a) = c/10^c to 10^c/log(a)=c/a,
you'll notice that 10^c is always larger than c and log(a) is always smaller than a (a larger than 1)...therefore 10^c/log(a) is always larger than c/a. So the determinant can never be zero. Same idea for (3).