If A = 3 -1 , 2 1 Show that A2 - 4 A + 5 I2 = 0. Also, find A4

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Let's first calculate $$A^2$$:

$A^2 = \begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} (3 \cdot 3 + (-1) \cdot 2) & (3 \cdot (-1) + (-1) \cdot 1) \\ (2 \cdot 3 + 1 \cdot 2) & (2 \cdot (-1) + 1 \cdot 1) \end{bmatrix}$

$A^2 = \begin{bmatrix} 9 - 2 & -3 - 1 \\ 6 + 2 & -2 + 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 7 & -4 \\ 8 & -1 \end{bmatrix}$

Now, let's calculate $$4A$$:

$4A = 4 \times \begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix}$

$4A = \begin{bmatrix} 12 & -4 \\ 8 & 4 \end{bmatrix}$

Finally, let's calculate $$5I_2$$ where $$I_2$$ is the 2x2 identity matrix:

$5I_2 = 5 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$5I_2 = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

Now, let's put it all together:

$A^2 - 4A + 5I_2 = \begin{bmatrix} 7 & -4 \\ 8 & -1 \end{bmatrix} - \begin{bmatrix} 12 & -4 \\ 8 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

$A^2 - 4A + 5I_2 = \begin{bmatrix} 7 - 12 + 5 & -4 + 4 + 0 \\ 8 - 8 + 0 & -1 - 4 + 5 \end{bmatrix}$

$A^2 - 4A + 5I_2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Therefore, $$A^2 - 4A + 5I_2 = 0$$.

Now, let's find $$A^4$$ by squaring $$A^2$$:

$A^4 = (A^2)^2$

$A^4 = \begin{bmatrix} 7 & -4 \\ 8 & -1 \end{bmatrix}^2$

Performing the matrix multiplication:

$A^4 = \begin{bmatrix} (7 \cdot 7 + (-4) \cdot 8) & (7 \cdot (-4) + (-4) \cdot (-1)) \\ (8 \cdot 7 + (-1) \cdot 8) & (8 \cdot (-4) + (-1) \cdot (-1)) \end{bmatrix}$

$A^4 = \begin{bmatrix} 49 - 32 & -28 + 4 \\ 56 - 8 & -32 + 1 \end{bmatrix}$

$A^4 = \begin{bmatrix} 17 & -24 \\ 48 & -31 \end{bmatrix}$

So, $$A^4 = \begin{bmatrix} 17 & -24 \\ 48 & -31 \end{bmatrix}$$.