Question: If A = $\begin{bmatrix} 1&4 \\ 2&3 \end{bmatrix}$. Find A$^{50}$
0

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

m4e(34) • 319 views
 modified 12 months ago  • written 16 months ago by
0

For characteristic equation |A - $\lambda$ I| = 0

$$\begin{vmatrix} 1 - \lambda & 4 \\ 2 & 3 - \lambda \end{vmatrix} = 0$$

$(1 - \lambda)(3 - \lambda) - 8 = 0 \\ 3 - \lambda - 3\lambda + \lambda^2 - 8 = 0 \\ \lambda^2 - 5\lambda + \lambda - 5 = 0 \\ (\lambda + 1)(\lambda - 5) = 0 \lambda = -1,5$

Let,

$\phi(A) = A^{50} \\ \phi(A) = aA + bI \\ \phi(\lambda) = \lambda^{50} \\ \phi(\lambda) = a\lambda + b \\ \phi(-1) = 1 \\ \phi(5) = 5^{50}$

case(i) $\lambda$ = -1

$\phi(-1) = a(-1) + b \\ a - b = 1 \hspace{0.25cm} ...(1)$

case(ii) $\lambda$ = 5

$\phi(5) = a(5) + b \\ 5^{50} = 5a + b \\ 5a + b = 5^{50} \hspace{0.25cm} ...(2) \\ 6a = 5^{50} - 1 \hspace{0.5cm} [From \,\, (1)] \\ a = \frac{5^{50} - 1}{6} \therefore b = \frac{5^{50} + 5}{6}$

$\phi(A) = aA + bI \\ = \frac{5^{50} - 1}{6} \begin{bmatrix} 1&4 \\ 2&3 \end{bmatrix} + \frac{5^{50} + 5}{6} \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \\ = \begin{bmatrix} \frac{5^{50} - 1}{6} & 4(\frac{5^{50} - 1}{6}) \\ 2(\frac{5^{50} - 1}{6}) & 3(\frac{5^{50} - 1}{6}) \end{bmatrix} + \begin{bmatrix} \frac{5^{50} + 5}{6} & 0 \\ 0 & \frac{5^{50} + 5}{6} \end{bmatrix} \\ = \begin{bmatrix} 2(\frac{5^{50} +2}{6}) & 4(\frac{5^{50} - 1}{6}) \\ 2(\frac{5^{50} - 1}{6}) & \frac{4 \times 5^{50} + 2}{6} \end{bmatrix} \\ = \begin{bmatrix} \frac{5^{50} + 2}{3} & 2(\frac{5^{50} - 1}{3}) \\ \frac{5^{50} - 1}{3} & \frac{2 \times 5^{50} + 1}{3} \end{bmatrix}$