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Math Help - matrix

  1. #1
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    matrix

    Say i have a matrix ,

    \begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5

    is it correct if i do it this way ,

    \begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5\begin{bmatrix}{1}&{0}\\{0}&  { 1}\end{bmatrix}


    =\begin{bmatrix}{9}&{3}\\{-1}&{12}\end{bmatrix}
    is 5 a scalar = 5I where I is an identity matrix ?
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Say i have a matrix ,

    \begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5 Mr F says: This makes no sense. Is this what the question said? Are you sure it didn't say {\color{red}\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5{\color{blue}I} } ....?

    is it correct if i do it this way ,

    \begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}+5\begin{bmatrix}{1}&{0}\\{0}&  { 1}\end{bmatrix}


    =\begin{bmatrix}{9}&{3}\\{-1}&{12}\end{bmatrix}
    is 5 a scalar = 5I where I is an identity matrix ?
    ..
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    ..
    GIven a matrix A=\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}

    And also given A^2-5A+6I=0 and i am supposed to deduce A^{-1} from this .

    so A(A-5)=-6I

    A(\frac{A-5}{-6})=I

    so A^{-1}=\frac{A-5}{-6}

    and my problem is here .. the (A-5) part . However , i did what i showed in my previous post and managed to get the answers .
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  4. #4
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    Why don't you just use the fact that

    If A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]

    then A^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]?
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Why don't you just use the fact that

    If A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]

    then A^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]?
    the question restricts me from doing that , it says hence deduce the inverse from that equation .
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  6. #6
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    Quote Originally Posted by thereddevils View Post
    GIven a matrix A=\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}

    And also given A^2-5A+6I=0 and i am supposed to deduce A^{-1} from this .

    so A(A-5)=-6I

    A(\frac{A-5}{-6})=I

    so A^{-1}=\frac{A-5}{-6}

    and my problem is here .. the (A-5) part . However , i did what i showed in my previous post and managed to get the answers .
    A^2-5A+6I=0

    Post multiply both sides by A^{-1}:

    A - 5I + 6A^{-1} = 0

    \Rightarrow A^{-1} = \frac{1}{6} (5I - A).
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