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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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    In reply to your attempt, you need to be sure that the elements of the diagonal of the resulting matrix do not multiply to zero as well. Remember this is one of the properties of the elements of $G$.2012-11-29

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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.2012-11-29
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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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    Maybe... it's always helped me personally though to think about the corresponding row/column vectors in matrices being multiplied2012-11-29