Question: If A = $\begin{bmatrix} 5&3 \\ 1&3 \end{bmatrix}$. Find A$^n$
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Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

m4e(34) • 338 views
 modified 12 months ago  • written 16 months ago by
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For characteristic equation |A - $\lambda$ I| = 0

$$\begin{vmatrix} 5 - \lambda & 3 \\ 1 & 3 - \lambda \end{vmatrix} = 0$$

$(5- \lambda)(3 - \lambda) - 3 = 0 \\ 15 - 5\lambda - 3\lambda + \lambda^2 - 3 = 0 \\ \lambda^2 - 8\lambda + 12 = 0 \\ (\lambda - 2)(\lambda - 6) = 0 \\ \lambda = 2,6$

Let,

$\phi(A) = A^n \\ \phi(A) = aA + bI \\ \phi(\lambda) = \lambda^n \\ \phi(\lambda) = a\lambda + b \\ \phi(2) = 2^n \\ \phi(6) = 6^n$

case(i) $\lambda$ = 2

$\lambda(2) = a(2) + b \\ 2^n = 2a + b \\ 2a + b = 2^n \hspace{0.25cm} ...(1)$

case(ii) $\lambda$ = 6

$\phi(6) = a(6) + b \\ 6^n = 6a + b \\ 6a + b = 6^n \hspace{0.25cm} ...(2)$

Solving (1) and (2) , we get,

$4a = 6^n - 2^n \\ a = \frac{6^n - 2^n}{4} \\ b = 2^n - 2a \\ = 2^n - 2(\frac{6^n - 2^n}{4}) \\ b = \frac{4 \times 2^n - 2 \times 6^n + 2 \times 2^n}{4} \\ = \frac{6 \times 2^n - 2 \times 6^n}{4}$

$\phi(A) = aA + bI \\ = \frac{6^n - 2^n}{4} \begin{bmatrix} 5&3 \\ 1&3 \end{bmatrix} + \frac{6 \times 2^n - 2 \times 6^n}{4} \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \\ = \begin{bmatrix} \frac{5 \times 6^n - 5 \times 2^n}{4} & \frac{3 \times 6^n - 3 \times 2^n}{4} \\ \frac{6^n - 2^n}{4} & \frac{3 \times 6^n - 3 \times 2^n}{4} \end{bmatrix} + \begin{bmatrix} \frac{6 \times 2^n - 2 \times 6^n}{4} & 0 \\ 0 & \frac{6 \times 2^n - 2 \times 6^n}{4} \end{bmatrix} \\ = \begin{bmatrix} \frac{3 \times 6^n - 2^n}{4} & \frac{3 \times 6^n - 3 \times 2^n}{4} \\ \frac{6^n - 2^n}{4} & \frac{6^n - 3 \times 2^n}{4} \end{bmatrix}$