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Thread: matrix

  1. #1
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    matrix

    Say i have a matrix ,

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

    is it correct if i do it this way ,

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


    $\displaystyle =\begin{bmatrix}{9}&{3}\\{-1}&{12}\end{bmatrix}$
    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 ,

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

    is it correct if i do it this way ,

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


    $\displaystyle =\begin{bmatrix}{9}&{3}\\{-1}&{12}\end{bmatrix}$
    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 $\displaystyle A=\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}$

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

    so $\displaystyle A(A-5)=-6I$

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

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

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

    then $\displaystyle A^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]$?
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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 $\displaystyle A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$

    then $\displaystyle A^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]$?
    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 $\displaystyle A=\begin{bmatrix}{4}&{3}\\{-1}&{7}\end{bmatrix}$

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

    so $\displaystyle A(A-5)=-6I$

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

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

    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 .
    $\displaystyle A^2-5A+6I=0$

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

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

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