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Thread: matrix products

  1. #1
    CT_Math
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    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CT_Math View Post
    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)
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  3. #3
    MHF Contributor red_dog's Avatar
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    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
    $\displaystyle A^{-1}(AB)=A^{-1}O_2\Leftrightarrow (A^{-1}A)B=O_2\Leftrightarrow I_2B=O_2\Leftrightarrow B=O_2$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    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
    $\displaystyle A^{-1}(AB)=A^{-1}O_2\Leftrightarrow (A^{-1}A)B=O_2\Leftrightarrow I_2B=O_2\Leftrightarrow B=O_2$
    that's a very elegant way to do it. i like it! my way was more brute-force-follow-your-nose
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    (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!
    $\displaystyle \left ( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ) \cdot \left ( \begin{array}{cc} 1 & -2 \\ -1 & 2 \end{array} \right ) = \left ( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right )$

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

    -Dan
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    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
    Yeah, we can create a more general formula to find such matrices when the determinant is zero
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    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
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