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:

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.

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.

• 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. Commented Dec 19, 2023 at 7:06
• @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. Commented Dec 19, 2023 at 7:24
• So 1 is false, while 2 is true! Commented Dec 19, 2023 at 7:49
• 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
Commented Dec 19, 2023 at 16:03
• What textbook are you using? I'd like to stay far away from it. Commented Dec 19, 2023 at 19:28

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

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.

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 $$β$$.

• Welcome to MSE. Please use MathJax to format your posts. Commented Dec 19, 2023 at 16:44
• Shall I edit? I'll rectify the MathJax or $\LaTeX$ one Commented Dec 19, 2023 at 17:43
• I was going to ask how one would "use MathJax" but it appears that I should simply use TeX math notation within my post, with nothing other than the dollar-sign delimiters to mark it. Thanks for fixing it up. Commented Dec 21, 2023 at 3:55

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!

• The 2nd image is the answer key, while I have typed down my working (thanks to @terran for editing the question). Commented Dec 19, 2023 at 7:50
• 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. Commented Dec 19, 2023 at 8:03
• I get your point: actually this is an app which trains us for competitives in India like JEE. Commented Dec 19, 2023 at 9:04
• There is a problem with the app question which I've reported to the developer. Commented Dec 19, 2023 at 9:05