Find the inverse of a matrix

**zorro** · December 28th 2009, 09:05 PM

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

**Prove It** · December 28th 2009, 09:24 PM

Originally Posted by zorro

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

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

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

**mr fantastic** · December 28th 2009, 10:44 PM

Originally Posted by zorro

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 $A^{-1}$ .

**zorro** · December 29th 2009, 12:18 AM

thanks mite

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