$A= \left[ \begin{array}{ccc} 3 & 10 & 5 \ -2 & -3 & -4 \ 3 & 5 & 7 \end{array}\right] $

Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks : 6 M

Year : Dec 2015

Characteristic equation is | A- λ I | =0

∴ $ \left| \begin{array}{ccc} 3-\lambda & 10 & 5 \\ -2 & -3-\lambda & -4 \\ 3 & 5 & 7-\lambda \end{array}\right|$

∴ (3- λ)[(-3-λ)(7-λ)+20]-10[-2( 7-λ)+12]+5[-10-3(-3-λ)]=0

∴ (3- λ)[ -21+3 λ-7λ+$λ^2$+20]-10[-14+2 λ+10] + 5[-10+9+3 λ]=0

∴ (3- λ) [$λ^2$-4 λ-1]-10[2λ-2]+5[3λ-1]=0

∴ $3[λ^2-12λ-3-λ^3+4λ^2+λ-20λ+20+15λ-5=0$

∴ $-λ^3+7λ^2-16λ+12=0$

∴$λ^3-7λ^2+16λ-12=0$

Cayley Hamilton theorem states that the characteristic equation is satisfied by A

∴ Replacing λ with A

∴$A^3-7A^2+16A-12=0$------(1)

Given : $A^6-6A^5+9 A^4+4 A^3-12 A^2+2A-I$-----(2)

Dividing equation (2) by equation (1) we get,

$ \hspace{4.2 cm} A^3+A^2 \\ A^3-7A^2+16A-12 ) \overline{ \; A^6-6A^5+9 A^4+4 A^3-12 A^2+2A-I }\\ \hspace{4 cm} A^6-7A^5+16A^4-12A^3 \\ \hspace{4 cm} - \; \; + \; \; \; \;- \; \; \; \; \; \ \ \ + \; \; \; \; \ \ \ \ \ \\ \hspace{3.5 cm} \overline{\hspace{1.2 cm} A^5-7 A^4+16 A^3-12 A^2+2A-I } \\ \hspace{4 cm} -A^5-7 A^4+16 A^3-12 A^2 \\ \hspace{4 cm} + \ \ \ - \ \ \ \ \ \ \ \ - \ \ \ \ \ \ + \\ \hspace{4 cm} \overline{\hspace{5 cm} 2A-I \hspace{2 cm}} \\ \; \\ \; \\ $

∴ Dividend = Divisor * Quotient + Remainder

∴ $A^6-6A^5+9 A^4+4 A^3-12 A^2+2A-I$ =$(A^3-7A^2+16A-12)*(A^3+ A^2 ) +(2A-1) \\ \;\\ = 0 * (A^3+ A^2 )+(2A-1) ----------(from (1)) \\ \; \\ = (2A-1)$

∴ $A^6-6A^5+9 A^4+4 A^3-12 A^2+2A-I=2A-1$