Let $R$ be a commutative ring with 1. Find squared matrixes $A_{ij}$ with $\det\biggl(\begin{bmatrix}A_{11} & A_{12}\\A_{21} & A_{22}\\\end{bmatrix}\biggl) \neq det(A_{11})\cdot det(A_{22}) - det(A_{12})\cdot det(A_{21}).$
My idea: I thought about inserting random numbers into a squared matrix but how do I know if $\det\biggl(\begin{bmatrix}A_{11} & A_{12}\\A_{21} & A_{22}\\\end{bmatrix}\biggl) \neq det(A_{11})\cdot det(A_{22}) - det(A_{12})\cdot det(A_{21}).$ actually applies?
Let's say that I have the following matrix: $\det\biggl(\begin{bmatrix}4 & 1\\2 & 3\\\end{bmatrix}\biggl) = 4\cdot3-1\cdot2=10.$
Do I get any valuable informations by doing this or what route should I take instead? I don't see how this: $\det\biggl(\begin{bmatrix}A_{11} & A_{12}\\A_{21} & A_{22}\\\end{bmatrix}\biggl) \neq det(A_{11})\cdot det(A_{22}) - det(A_{12})\cdot det(A_{21})$ is even possible at all.
Any hints guding me to the right direction I much appreciate.
Let \begin{eqnarray*} A_{11}=A_{22}= \begin{bmatrix} 1 &1 \\1 &1 \\ \end{bmatrix} \end{eqnarray*} and \begin{eqnarray*} A_{12}=A_{12}= \begin{bmatrix} 1 &1 \\1 &2 \\ \end{bmatrix} . \end{eqnarray*} So \begin{eqnarray*} A= \begin{bmatrix} 1 &1 &1&1 \\1 &1 &1&2\\1 &1 &1&1 \\1 &2 &1&1\\ \end{bmatrix} . \end{eqnarray*} Now calculate the determinants.