Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 6M

Year: May 2015

Characteristic equation of A is

$\begin{bmatrix} 7 - \lambda & 4 & -1 \\ 4 & 7 - \lambda & -1 \\ -4 & -4 & 4 -\lambda \end{bmatrix} = 0$

Simplifying we get

$\therefore (7 - \lambda) [(7 - \lambda) (4 - \lambda) - 4] - 4[4(4 - \lambda) - 4] - 1[-16 + 4(7 - \lambda)] = 0 \\ \therefore (7 - \lambda) [28 - 4\lambda - 7\lambda + \lambda^2 - 4] - 4[16 - 4\lambda - 4] - 1[12 - 4\lambda] = 0 \\ \therefore (7 - \lambda) [ \lambda^2 - 11\lambda + 24] - 4[12 - 4\lambda] + 4\lambda - 12 = 0 \\ \therefore 7\lambda^2 - 77\lambda + 168 - \lambda^3 + 11\lambda^2 - 24\lambda - 48 + 16\lambda + 4\lambda - 12 = 0 \\ \therefore -\lambda^3 + 18\lambda^2 - 81\lambda + 108 = 0$

Solving it we get

$\lambda = 12, 3, 3$

Hence the roots of $|A - \lambda I| = 0$ are 3,3, 12

Let us now find the minimal polynomial of A we know that each characteristics root of A is also a root of the minimal polynomial of A so if f(x) is the minimal polynomial of A then (x - 3) & (x - 12) are the factors of f(x) let us see whether (x - 3)(x - 12) = $x^2 - 15x + 36$ annihilates A.

Now

$A^2 = \begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & -4 \end{bmatrix}^2 \\ = \begin{bmatrix}7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & -4 \end{bmatrix} \begin{bmatrix}7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & -4 \end{bmatrix} \\ = \begin{bmatrix} 69 & 60 & -15 \\ 60 & 69 & -15 \\ -60 & -60 & 24 \end{bmatrix} \\ \therefore A^2 - 15A + 36I = \begin{bmatrix} 69 & 60 & -15 \\ 60 & 69 & -15 \\ -60 & -60 & 24 \end{bmatrix} - 15 \begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & -4 \end{bmatrix} + 36 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \therefore f(x) = x^2 - 15x + 36$

annihilates A