# Find the inverse of a matrix

• Dec 28th 2009, 10:05 PM
zorro
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????
• Dec 28th 2009, 10:24 PM
Prove It
Quote:

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

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

$= \left[\begin{matrix} -\frac{5}{2} & \frac{3}{2}\\ 2 & -1\end{matrix}\right]$.
• Dec 28th 2009, 11:44 PM
mr fantastic
Quote:

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 $A^{-1}$.
• Dec 29th 2009, 01:18 AM
zorro
thanks mite (Beer)