Let \[ M=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right) \quad \text{and} \quad N=\left(\begin{array}{cc}{x} & {u} \\ {-y} & {-v}\end{array}\right) \]
over a commutative ring \(R\), where \(b\) and \(x\) are units of \(R\). Prove that \[ M N=\left(\begin{array}{ll}{0} & {0} \\ {0} & {*}\end{array}\right) \implies MN = 0 .\]
Solution
Multiply everything out to get \[ \matt{ax-by}{au-bv}{cx-dy}{cu-dv} ,\] so it suffices to show \(cu=dv\) given \[ ax &= by \\ cx &= dy \\ au &= bv .\]
Writing \(cu\):
Use that \(b\in \unitsof{R}\), left-multiply (1) by \(\inverseof{b}\) to get \(\inverseof{b} a x = y\)
Substitute \(y\) into (2) to get \(cx = d(\inverseof{b} a x)\).
Since \(x\in \unitsof{R}\), right-multiply by \(\inverseof{x}\) to get \(c = d\inverseof{b} a\) and thus \(cu = d\inverseof{b} a u\).
Summary: \[ ax = by &\implies \inverseof{b} ax = y \\ &\implies cx = dy = d(\inverseof{b} a x) \\ &\implies c = d\inverseof{b} a \\ &\implies cu = d\inverseof{b} au .\]
Writing \(dv\):
Left-multiply (3) by \(\inverseof{b}\) to get \(\inverseof{b} au = v\).
Left-multiply by \(d\) to get \(d\inverseof{b} au = dv\)
Summary: \[ au = bv &\implies \inverseof{b} a u = v \\ &\implies d\inverseof{b} au = dv .\]
So \[ cu = d\inverseof{b} a u = dv .\]