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Math Help - Find the inverse of a matrix

  1. #1
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    Find the inverse of a matrix

    If A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} then verify that A^2 - 7 A - 2I = 0 . Also find A^{-1}
    ---------------------------------------------------------------------
    Solution :

    A^2 = \begin{bmatrix}16 & 21 \\ 28 & 37 \end{bmatrix}....................Is this correct ????

    A^2 - 7 A - 2I = 0 = 0 ...............Is this correct ????

    A{-1} = \frac{1}{|A|} . Adj(Co factor of A)

    \therefore Adj(Co factor of A) = \begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix}.........................Is this correct ????

    |A| = -2


    \therefore A^{-1}  = \begin{bmatrix} \frac{5}{2} & \frac{3}{2 } \\2 & -1 \end{bmatrix}...................Is this correct????
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  2. #2
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    Quote Originally Posted by zorro View Post
    If A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} then verify that A^2 - 7 A - 2I = 0 . Also find A^{-1}
    ---------------------------------------------------------------------
    Solution :

    A^2 = \begin{bmatrix}16 & 21 \\ 28 & 37 \end{bmatrix}....................Is this correct ????

    A^2 - 7 A - 2I = 0 = 0 ...............Is this correct ????

    A{-1} = \frac{1}{|A|} . Adj(Co factor of A)

    \therefore Adj(Co factor of A) = \begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix}.........................Is this correct ????

    |A| = -2


    \therefore A^{-1} = \begin{bmatrix} \frac{5}{2} & \frac{3}{2 } \\2 & -1 \end{bmatrix}...................Is this correct????
    A^2 is correct.

    You haven't shown the equation is true yet though...

    A^2 - 7A - 2I = \left[\begin{matrix}16 & 21\\ 28 & 37 \end{matrix}\right] - 7\left[\begin{matrix} 2 & 3\\ 4 & 5 \end{matrix}\right] - 2\left[\begin{matrix}1 & 0\\ 0 & 1\end{matrix}\right]

     = \left[\begin{matrix}16 & 21\\ 28 & 37 \end{matrix}\right] - \left[\begin{matrix} 14 & 21\\ 28 & 35 \end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]

     = \left[\begin{matrix} 0 & 0\\ 0& 0\end{matrix}\right]

     = \mathbf{0}.


    For a 2 \times 2 matrix \left[\begin{matrix} a & b \\ c & d\end{matrix}\right], the inverse is

    \frac{1}{ad - bc}\left[\begin{matrix}d & -b\\ -c & a\end{matrix}\right].

    So A^{-1} = \frac{1}{2 \cdot 5 - 3 \cdot 4}\left[\begin{matrix} 5 & -3\\ -4 & 2\end{matrix}\right]

     = -\frac{1}{2}\left[\begin{matrix} 5 & -3\\ -4 & 2\end{matrix}\right]

     = \left[\begin{matrix} -\frac{5}{2} & \frac{3}{2}\\ 2 & -1\end{matrix}\right].
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  3. #3
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    Quote Originally Posted by zorro View Post
    If A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} then verify that A^2 - 7 A - 2I = 0 . Also find A^{-1}
    ---------------------------------------------------------------------
    Solution :

    A^2 = \begin{bmatrix}16 & 21 \\ 28 & 37 \end{bmatrix}....................Is this correct ????

    A^2 - 7 A - 2I = 0 = 0 ...............Is this correct ????

    A{-1} = \frac{1}{|A|} . Adj(Co factor of A)

    \therefore Adj(Co factor of A) = \begin{bmatrix} 5 & -3 \\ -4 & 2 \end{bmatrix}.........................Is this correct ????

    |A| = -2


    \therefore A^{-1} = \begin{bmatrix} \frac{5}{2} & \frac{3}{2 } \\2 & -1 \end{bmatrix}...................Is this correct????
    You have A^2 - 7 A - 2I = 0. Left multiply by A^{-1}: A - 7I - 2A^{-1} = 0. Re-arrange and solve for A^{-1}.
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  4. #4
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    thanks mite
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