How would I find a 2x2 matrix to multiply another 2x2 matrix by to get zero if that matrix is not the zero matrix? Specifically I have
[1 2]
[3 4] times B = 0
Let the unknown matrix be $\displaystyle B = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$
We want:
$\displaystyle \left(\begin{array}{cc}1&2\\3&4\end{array}\right) \left(\begin{array}{cc}a&b\\c&d\end{array}\right) = \left(\begin{array}{cc}0&0\\0&0\end{array}\right)$
$\displaystyle \Rightarrow \left(\begin{array}{cc} a + 2c & b + 2d \\ 3a + 4c & 3b + 4d \end{array}\right) = \left(\begin{array}{cc}0&0\\0&0\end{array}\right)$
Thus we need the solution to the systems:
$\displaystyle a + 2c = 0$ ..............(1)
$\displaystyle 3a + 4c = 0$ ..............(2)
and
$\displaystyle b + 2d = 0$ ....................(1)
$\displaystyle 3b + 4d = 0$ ...................(2)
Can you take it from here?
(That's the systematic way to do it, but really, as with numbers, multiplying any number by zero gives zero, so too with matrices: any matrix times the zero matrix gives the zero matrix, you will find that the solution to the above systems is a = b = c = d = 0)
Let $\displaystyle \displaystyle A=\left(\begin{array}{cc}1 & 2\\3 & 4\end{array}\right)$. We have $\displaystyle \det A=-2$, so $\displaystyle A$ is invertible. Then multiplying both members at left with $\displaystyle A^{-1}$, we have
$\displaystyle A^{-1}(AB)=A^{-1}O_2\Leftrightarrow (A^{-1}A)B=O_2\Leftrightarrow I_2B=O_2\Leftrightarrow B=O_2$
However beware!
$\displaystyle \left ( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ) \cdot \left ( \begin{array}{cc} 1 & -2 \\ -1 & 2 \end{array} \right ) = \left ( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right )$
(This is due to red_dog's condition $\displaystyle \text{det}A \neq 0$, which is violated here.)
-Dan