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Thread: Find the inverse of a matrix

  1. #1
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    Find the inverse of a matrix

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

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

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

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

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

    $\displaystyle |A| = -2$


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

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

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

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

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

    $\displaystyle |A| = -2$


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

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

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

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

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

    $\displaystyle = \mathbf{0}$.


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

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

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

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

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

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

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

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

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

    $\displaystyle |A| = -2$


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