Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

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

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

$ (3 -\lambda)(3 -\lambda) - 1 = 0 9 -6 \lambda + \lambda^2 - 1 = 0 \lambda^2 - 4 \lambda - 2\lambda + 8 = 0 \\ \lambda(\lambda-4) - 2(\lambda-4) = 0 \\ (\lambda-2)(\lambda-4) = 0 \lambda = 2,4 $

Let,

$ \phi (A) = 5^A \\ \phi(A) = aA + bI \\ \phi (\lambda) = 5^{\lambda} \\ \phi (\lambda) = a \lambda + b \\ \phi(2) = 5^2 = 25 \\ \phi(4) = 5^4 = 625 $

case(i): $\lambda$ = 2

$ \phi(2) = a(2) + b \\ 25 = 2a + b \\ 2a + b = 25 $

case(ii): $\lambda$ = 4

$ \phi(4) = a(4) + b \\ 5^4 = 4a + b \\ 4a + b = 625 $

Solving both the cases, we get, a = 300 and b = -575

$ \phi(A) = aA + bI = 300 \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix} + (-575) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} 900 & 300 \\ 300 & 900 \end{bmatrix} - \begin{bmatrix} 575 & 0 \\ 0 & 575 \end{bmatrix} \\ = \begin{bmatrix} 325 & 300 \\ 300 & 325 \end{bmatrix} \\ \therefore 5^A = \begin{bmatrix} 325 & 300 \\ 300 & 325 \end{bmatrix} $