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If A = $\begin{bmatrix} 2& 3 \\ -3&-4 \end{bmatrix}$. Prove that A$^{50}$ = $\begin{bmatrix} -149&150 \\ 150&151 \end{bmatrix}$

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

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

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

$ (2 - \lambda)(-4- \lambda) + 9 = 0 \\ (\lambda -2)(\lambda+4) + 9 = 0 \\ \lambda^2 + 2\lambda -8+9 = 0 \\ \lambda^2 + 2\lambda +1 = 0 \\ (\lambda + 1)^2 = 0 \\ \lambda = -1,-1 $

Let,

$ \phi(A) = A^{50} \\ = aA = bI \\ \phi(\lambda) = \lambda^{50} \\ = a\lambda + b \hspace{0.25cm} ...(1) \\ \phi(-1) = 1 $

case(i) $\lambda$ = -1

$ \phi(-1) = a(-1) + b \\ 1 = -a + b \\ a - b = -1 $

case(ii) $\lambda$ = -1

Since the eigen values are repeated, differentiate (1) w.r.t.$\lambda$

$ \phi'(\lambda) = a(1) \\ -50 = a \\ \implies a = -50 \\ \phi'(\lambda) = 50\lambda^{49} \\ = 50(-1)^{49} = -50 $

$ -50 -b = -1 \implies b = -49 $

$ \phi(A) = aA + bI \\ A^{50} = -50A - 49I \\ = -50 \begin{bmatrix} 2 & 3 \\ -3 & -4 \end{bmatrix} - 49 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} -100 & -150 \\ 150 & 200 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} -149 & -150 \\ 150 & 151 \end{bmatrix} $

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