# Show that if matrices A and B are elements of G, then AB is also an element of G.

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?

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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

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 $AB = P \in G$.

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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.

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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.

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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