Answer to Question #153071 in Linear Algebra for Kishore Kris

Question #153071
2. ſi 0 Verify Cayley-Hamilton theorem, Find A ^ 4 when A= 2 1 1 1 1 -1 3 -1 1
1
Expert's answer
2020-12-29T16:01:37-0500

A=[211111311]A=\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}

Characteristic polynomial of A,

det(AλI)=0det(A-\lambda I)=0

det([211111311]λ[100010001])=0\Rightarrow det\left(\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix} - \lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\right)=0

det([2λ1111λ1311λ])=0\Rightarrow det \left(\begin{bmatrix}2-\lambda&1&1\\1&1-\lambda&-1\\3&-1&1-\lambda\end{bmatrix}\right)=0

(2λ)det([1λ111λ])\Rightarrow (2-\lambda)det\left(\begin{bmatrix}1-\lambda&-1\\-1&1-\lambda\end{bmatrix}\right)

det([1131λ])-det\left(\begin{bmatrix}1&-1\\3&1-\lambda\end{bmatrix}\right)

+det([11λ31])=0+det\left(\begin{bmatrix}1&1-\lambda\\3&-1\end{bmatrix}\right)=0


(2λ)(λ22λ)(λ+4)+(3λ4)=0\Rightarrow \left(2-\lambda\right)\left(\lambda^2-2\lambda\right)-\left(-\lambda+4\right)+ \left(3\lambda-4\right)=0

λ34λ2+8=0\Rightarrow \lambda^3-4\lambda^2+8=0

\therefore Characteristic polynomial of A :- λ34λ2+8=0\lambda^3-4\lambda^2+8=0


Now,

A2=A×A=[211111311]×[211111311]A^2= A\times A =\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}\times \begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}


=[822031815]=\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}

and,

A3=A2×A=[822031815]×[211111311]A^3=A^2\times A=\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}\times \begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}


=[248804432412]=\begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}


Now, by Cayley-Hamilton theorem, putting λ=A\lambda=A ,

A34A2+8I=03×3A^3-4A^2+8I=0_{3\times3}

[248804432412]4[822031815]\Rightarrow \begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}-4\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}


+8[100010001]=03×3+8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=0_{3\times3}


[000000000]=03×3\Rightarrow\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0_{3\times3}


\therefore Cayley-Hamilton Theorem Verified.


Now, by Cayley-Hamilton theorem,


A34A2+8I=03×3A^3-4A^2+8I=0_{3\times3}

A(A34A2+8I)=A×03×3\Rightarrow A(A^3-4A^2+8I)=A\times0_{3\times3}

[ Multiplying both side by matrix AA ]

A44A3+8A=03×3\Rightarrow A^4-4A^3+8A=0_{3\times3}

A4=4A38A\Rightarrow A^4 = 4A^3-8A

=4[248804432412]8[211111311]=4\begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}-8\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix} =[8024248881042440]=\begin{bmatrix}80&24&24\\-8&8&-8\\104&24&40\end{bmatrix}



A4=[8024248881042440]\therefore A^4=\begin{bmatrix}80&24&24\\-8&8&-8\\104&24&40\end{bmatrix}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment