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I tried the following problem.

Let $A = \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $B = \begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$.

There are 2 statements:

  1. $AB - BA$ is always an invertible matrix, and
  2. $AB - BA$ is never an identity matrix.

Now I'm being asked to find out whether these statements are true or false.

I'm not too familiar with $\LaTeX$ and I don't have enough rep to upload images, I had to put my problem as a linked image:

MY PROBLEM

On calculation:

I first got the product of the two matrices as:

$AB = \begin{bmatrix} 0 & \alpha\gamma \\ \beta\delta & 0 \end{bmatrix}$

$BA = \begin{bmatrix} 0 & \beta\gamma \\ \alpha\delta & 0 \end{bmatrix}$

$AB - BA = \begin{bmatrix} 0 & (\alpha-\beta)\gamma \\ (\beta-\alpha)\delta & 0 \end{bmatrix}$

From the above I got that $AB-BA$ cannot be an identity matrix since the main diagonal elements are all zero. So statement 2 is correct, I believe.

And I also felt Statement 1 too was correct, since $|AB-BA|=\gamma\delta(\alpha-\beta)^2$. (Ouchie, it was quite careless of me!)

But the answer tells me a different story.

ANSWER KEY

I deduced that there are two possibilities: either the answer given is incorrect, or I have made an error somewhere and I am not able to identify.

Xander Henderson
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Harikrishnan M
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    Since the question does not specify the values of $\alpha,\beta,\gamma$ and $\delta$, there is no way to judge whether statement 1 is true or not. All we can say is that $AB-BA$ is invertible if and only if $\alpha-\beta,,\gamma$ and $\delta$ are nonzero. E.g. it is invertible when $\alpha=\gamma=\delta=1$ and $\beta=0$, but it is non-invertible when $\alpha=\beta=\gamma=\delta=0$. Statement 2 is true because $AB-BA$ has a zero diagonal. Your given answers for both statements are wrong. – user1551 Dec 19 '23 at 07:06
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    @user1551 The original statement 1 (as seen in the image) is "$AB-BA$ is always an invertible matrix". That has a definite truth value, assuming something reasonable, like $\alpha, \beta, \gamma, \delta$ being arbitrary real numbers. – Arthur Dec 19 '23 at 07:24
  • So 1 is false, while 2 is true! – Harikrishnan M Dec 19 '23 at 07:49
  • $|AB-BA|=-\gamma\delta(\alpha-\beta)^2$ when $\gamma = 0$ or when $\delta=0$ or when $\alpha=\beta$. Indeed when $\gamma = 0$ or $\delta=0$ you get at least three zeros in $AB-BA$, while when $\alpha=\beta$ you get all four being zero. – Henry Dec 19 '23 at 15:04
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    Note that for statement 1 there is the word "always" in the image but not in the text at the beginning of your post. This is an important word, because otherwise there are no quantifiers over $\alpha, \beta, \gamma, \delta$, so they appear to be free variables, and it's unclear what statement 1 is saying without this "always". – Stef Dec 19 '23 at 16:03
  • Your $\cal{L}$ATE$X$ is good. – Bob Dobbs Dec 19 '23 at 16:38
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    What textbook are you using? I'd like to stay far away from it. – Mad Physicist Dec 19 '23 at 19:28
  • Well it is not a textbook. It's an app used to practice for competitives like JEE Main and Advanced at my place – Harikrishnan M Dec 20 '23 at 00:59
  • It's called MARKS and it does have quite good questions but ultimately outdated questions too from 2000's. And this is from 2012. – Harikrishnan M Dec 20 '23 at 01:05

3 Answers3

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"I deduced that there are two possibilities" : Actually , there are three Possibilities !

Here , answer given is incorrect & OP made mistakes too.

OP is almost right that $AB-BA$ has Determinant $\color{blue}{+}\gamma\delta(\alpha-\beta)^2$ (( the Outer Symbol is $\color{blue}{+}$ , not $\color{red}{-}$ , thanks to new user "kevin martin" who caught that ))
[[ Text Book is wrong that Determinant is $(\alpha-\beta)^2\delta$ ]]

When either $\gamma=0$ or $\delta=0$ or $\alpha=\beta$ , Determinant is $0$ & that Matrix will have no Inverse.
Hence Statement 1 is not true.

OP is right that $AB-BA$ is not $I$.
Text Book is wrong that $AB-BA$ can somehow become $I$
It is never Identity Matrix.
Hence Statement 2 is true.
[[ It might be "Symmetric Matrix" , when $\delta=-\gamma$ ]]

OBSERVATIONS :

Text Book has a lot of typos :
Eg misplaced $)]$ to $])$
Eg mispaced $(\alpha-\beta)^2$ to $(\alpha-\beta^2\rangle$

Text Book has a lot of Assumptions :
Eg Non-Zero Elements
Eg Non-Equal Elements

Text Book is wrong in Basics :
Eg Mixing Identity with Symmetry
Eg Writing $\gamma=-\delta$ AND $\delta=-\gamma$ which is repeating SAME CRITERIA !

SUGGESTION : Use better text book to Improve yourself.

Check out :
https://brilliant.org/wiki/matrices/
https://brilliant.org/courses/linear-algebra/
https://openstax.org/details/books/college-algebra-2e
https://open.umn.edu/opentextbooks/textbooks/213
https://en.wikibooks.org/wiki/Introductory_Linear_Algebra/Matrices
https://en.wikibooks.org/wiki/Introductory_Linear_Algebra/Matrix_inverses_and_determinants

Prem
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I think that OP and the reference both have the determinant wrong. I find that the correct determinant is: $|AB−BA|=γδ(α−β)^2.$ From the standard formula for a $2×2$ matrix, the determinant is $-γ(α−β))(δ(β−α))$. The first negation comes from the formula. This is canceled by the fact that the difference between $α$ and $β$ is reversed in one of the values.

As long as $γ≠0$ and $δ≠0$ and $α≠β$ then $|AB−BA|$ will be invertible. Otherwise the determinant is zero and $|AB−BA|$ is not invertible.

The book is also wrong about being an identity matrix. Because the main diagonal is not all 1's this can never be an identity matrix, regardless of the values of $γ$, $δ$, $α$, and $β$.

Harikrishnan M
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Kevin Martin
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A small error in both. First, the Identity matrix $\mathbb{I}$ has 1 on the diagonal going down from the upper left corner, and 0 everywhere else. It does NOT have 1 on the diagonal from the other corner.

$$\begin{pmatrix} 1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 \end{pmatrix}$$ Is how an identity matrix should look. Notice which diagonal has $1$s on it. We also know that $\mathbb{I}A = A$ for the identity matrix and any equally sized A, so if you’re ever on a test and in trouble I would recommend trying this with some $3x3$ matrix to check yourself.

On your statement 1, you have all of the e work done correctly! However, if we set $\delta = 0$, or some other way of getting 0 given those values, our determinant is $0$ and hence we are non invertible (or singular if you’re feeling fancy).

These are small definition based mistakes, so keep at it and you will get better! Also good post formatting, it was clear and easy to read, but you may want to make more clear whether the second image is your work or the text answer (if it is the text get a new one). Welcome to MSE!

Robertmg
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  • The 2nd image is the answer key, while I have typed down my working (thanks to @terran for editing the question). – Harikrishnan M Dec 19 '23 at 07:50
  • Go get a new book then. No textbook should make those mistakes. I recommend contemporary linear algebra by Anton Busby, as they sell hardcover copies for under $10$$ on thriftbooks, and it is also a very good book. – Robertmg Dec 19 '23 at 08:03
  • I get your point: actually this is an app which trains us for competitives in India like JEE. – Harikrishnan M Dec 19 '23 at 09:04
  • There is a problem with the app question which I've reported to the developer. – Harikrishnan M Dec 19 '23 at 09:05