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

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- Jul 7th 2007, 06:39 AMCT_Mathmatrix products
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 - Jul 7th 2007, 06:50 AMJhevon
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) - Jul 7th 2007, 08:17 AMred_dog
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

$\displaystyle A^{-1}(AB)=A^{-1}O_2\Leftrightarrow (A^{-1}A)B=O_2\Leftrightarrow I_2B=O_2\Leftrightarrow B=O_2$ - Jul 7th 2007, 08:18 AMJhevon
- Jul 7th 2007, 02:11 PMtopsquark
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 - Jul 7th 2007, 02:30 PMJhevon
- Jul 7th 2007, 04:21 PMtopsquark