# matrix products

• Jul 7th 2007, 07:39 AM
CT_Math
matrix 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, 07:50 AM
Jhevon
Quote:

Originally Posted by CT_Math
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 $B = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$

We want:

$\left(\begin{array}{cc}1&2\\3&4\end{array}\right) \left(\begin{array}{cc}a&b\\c&d\end{array}\right) = \left(\begin{array}{cc}0&0\\0&0\end{array}\right)$

$\Rightarrow \left(\begin{array}{cc} a + 2c & b + 2d \\ 3a + 4c & 3b + 4d \end{array}\right) = \left(\begin{array}{cc}0&0\\0&0\end{array}\right)$

Thus we need the solution to the systems:

$a + 2c = 0$ ..............(1)
$3a + 4c = 0$ ..............(2)

and

$b + 2d = 0$ ....................(1)
$3b + 4d = 0$ ...................(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, 09:17 AM
red_dog
Let $\displaystyle A=\left(\begin{array}{cc}1 & 2\\3 & 4\end{array}\right)$. We have $\det A=-2$, so $A$ is invertible. Then multiplying both members at left with $A^{-1}$, we have
$A^{-1}(AB)=A^{-1}O_2\Leftrightarrow (A^{-1}A)B=O_2\Leftrightarrow I_2B=O_2\Leftrightarrow B=O_2$
• Jul 7th 2007, 09:18 AM
Jhevon
Quote:

Originally Posted by red_dog
Let $\displaystyle A=\left(\begin{array}{cc}1 & 2\\3 & 4\end{array}\right)$. We have $\det A=-2$, so $A$ is invertible. Then multiplying both members at left with $A^{-1}$, we have
$A^{-1}(AB)=A^{-1}O_2\Leftrightarrow (A^{-1}A)B=O_2\Leftrightarrow I_2B=O_2\Leftrightarrow B=O_2$

that's a very elegant way to do it. i like it! my way was more brute-force-follow-your-nose
• Jul 7th 2007, 03:11 PM
topsquark
Quote:

Originally Posted by Jhevon
(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)

However beware!
$\left ( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ) \cdot \left ( \begin{array}{cc} 1 & -2 \\ -1 & 2 \end{array} \right ) = \left ( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right )$

(This is due to red_dog's condition $\text{det}A \neq 0$, which is violated here.)

-Dan
• Jul 7th 2007, 03:30 PM
Jhevon
Quote:

Originally Posted by topsquark
However beware!
$\left ( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ) \cdot \left ( \begin{array}{cc} 1 & -2 \\ -1 & 2 \end{array} \right ) = \left ( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right )$

(This is due to red_dog's condition $\text{det}A \neq 0$, which is violated here.)

-Dan

Yeah, we can create a more general formula to find such matrices when the determinant is zero
• Jul 7th 2007, 05:21 PM
topsquark
Quote:

Originally Posted by Jhevon
Yeah, we can create a more general formula to find such matrices when the determinant is zero

Which pretty much can only be done by the method that you posted. ;)

-Dan