Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be the set of $2 \times 2$ matrices of the form \begin{pmatrix} a & b \\ 0 & c\end{pmatrix} such that $ac$ is not zero. Show that if matrices $A$ and $B$ are elements of $G$, then $AB$ is also an element of $G$.

Do I just need to show that $AB$ has a non-zero determinant?

share|cite|improve this question
What have you tried? Hint: pick two matrices of the given form and multiply them. – Code-Guru Nov 29 '12 at 21:05
That's what I did; I had A=a,b,c,0 (clockwise) and B=d,e,f,0 (clockwise) and multiplied them together. The determinent was then adcf and as ac is not zero, I said AB is an element of G if df is also not 0. – Mathlete Nov 29 '12 at 21:06
Please edit your question so that you can provide the correct formatting. – Code-Guru Nov 29 '12 at 21:09
I'm afraid I haven't gotten to grips with LaTex yet... – Mathlete Nov 29 '12 at 21:09
And LaTex for more than the simplest expressions isn't that great in a comment. Feel free to edit your original question. – Code-Guru Nov 29 '12 at 21:10
up vote 6 down vote accepted

Let $$A = \begin{bmatrix} a & b \\ 0 & c\\ \end{bmatrix}, \;\; B = \begin{bmatrix} e & f \\ 0 &g \\ \end{bmatrix}$$

where $ac\neq 0,\;\;eg \neq 0$. So $A, B \in G$.

Simply compute $AB= P$ and what to you get? Use the definition of matrix multiplication, and the fact that $ac \neq 0$ and $eg\neq 0$, and check to see if the lower left entry of your product matrix $P$ is, in fact, $0$.

Showing that $\det (AB) = \det(P) \neq 0$ is not your task. In fact, the $$\det \begin{bmatrix} m & 0\\n& q\\ \end{bmatrix} \neq 0$$ when $m, n, q$ are non-zero, but this matrix is NOT in $G$.

You need to verify that for the entries $p_{ij}$ of $AB = P$:

$p_{11}p_{22} \neq 0.$

$p_{21} = 0$.

Once you've done that, you can conclude $AB = P \in G$.

share|cite|improve this answer
That's what I did but I didn't think to state that eg is non-zero first. Makes a lot more sense, thanks. – Mathlete Nov 29 '12 at 21:10

Proving that AB has a non-zero determinant is not enough, because not all 2x2 matrices with non-zero determinant are a element of G.

You need to prove another property of AB. This property is that it has the shape you stated.

This combined with a non-zero determinant guarantees that AB has the prescribed shape with ac not zero.

share|cite|improve this answer

Another way to put it: entry $G_{21}$ is given by the dot product of vectors $(0 \space A_{22})$ and $(B_{11} \space 0)$. These are orthogonal, ie their dot product is zero, so that entry is always 0.

share|cite|improve this answer
Too difficult for somebody asking elementary things about matrices I think. – Applied mathematician Nov 29 '12 at 21:16
Maybe... it's always helped me personally though to think about the corresponding row/column vectors in matrices being multiplied – tacos_tacos_tacos Nov 29 '12 at 21:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.