Subject: Applied Mathematics 2

Topic: Matrices

Difficulty: High

Find the characteristic equation of the matrix A given below and hence ,find the matrix represented by $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$ , where $A = \begin{bmatrix} \ 2 & 1 & 1 \\ \ 0 & 1 & 0 \\ \ 1 & 1 & 2 \\ \end{bmatrix}$

Characteristic eqn |A- λI| =0

i.e. $ \begin{vmatrix} 1-λ&2&-2 \\ -1&3-λ&0 \\ 0&-2&1-λ \end{vmatrix} = 0 $

Expanding we get a cubic equation in λ, as $ λ^3 - 5λ^2 + 7λ - 3 =0$

By C-H theorem we have $A^3 - 5A^2 + 7A – 3I =0$ ……….(1)

To find the value of given matrix eqn , divide the matrix eqn

$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$ by the L.H.S. of eqn (1),

we get quotient $A^5+A$ & the remainder $A^2+A+I$ ,it can be written as

$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I $

= $( A^3 - 5A^2 + 7A – 3I ) (A^5+A) + (A^2+A+I ) $

$= A^2+A+I \,\,\,\,\,\,\, $ ( using eqn 1)

$ = \begin{bmatrix} 5&4&4 \\ 0&1&0 \\ 4&4&5 \end{bmatrix} + \begin{bmatrix} 2&1&1 \\ 0&1&0 \\ 1&1&2 \end{bmatrix} + \begin{bmatrix} 1& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix} 8& 5& 5 \\ 0& 3& 0 \\ 5& 5& 8 \end{bmatrix} $