The Cayley-Hamilton Theorem provides a method for calculating powers of a matrix. For example, if $ A $ is a $ 3 \times 3 $ matrix with the characteristic equation\[ c_0 + c_1 \lambda + c_2 \lambda^2 + \lambda^3 = 0 \]then $ c_0I + c_1A + c_2A^2 + A^3 = 0 $, so\[ A^3 = -c_2A^2 - c_1A - c_0I \]Multiplying through by $ A $ yields \[ A^4 = -c_0A - c_1A^2 - c_2A^3 \]which expresses $ A^4 $ in terms of $ A^3, A^2 $, and $ A $. Use this procedure to calculate $ A^3 $ and $ A^4 $ for \[ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -4 & 4 \end{bmatrix} \]$ A^3 $ is\[\begin{bmatrix} \boxed{} & \boxed{} & \boxed{} \\ \boxed{} & \boxed{} & \boxed{} \\ \boxed{} & \boxed{} & \boxed{} \end{bmatrix}\]$ A^4 $ is\[\begin{bmatrix} \boxed{} & \boxed{} & \boxed{} \\ \boxed{} & \boxed{} & \boxed{} \\ \boxed{} & \boxed{} & \boxed{} \end{bmatrix}\]

The matrix is A=0100011-44. Obtain the characteristic equation:…

The Cayley-Hamilton Theorem provides a method for calculating powers of a matrix. For example, if A is a 3x 3 matrix with the characteristic equation Co + Cz1 +C212 + 13 = 0 then col + c,A + C2A² + A³ = 0, so A3 = - C2A? - C1A – col Multiplying through by A yields A4 = - coA – c¿A² – C2A³, which expresses A“ in terms of A3, A² and A. Use this procedure to calculate A3 and A4 for 0. 1 0 A =| 0 0 1 1 -4 4

The Cayley-Hamilton Theorem provides a method for calculating powers of a matrix. For example, if A is a 3x 3 matrix with the characteristic equation Co + Cz1 +C212 + 13 = 0 then col + c,A + C2A² + A³ = 0, so A3 = - C2A? - C1A – col Multiplying through by A yields A4 = - coA – c¿A² – C2A³, which expresses A“ in terms of A3, A² and A. Use this procedure to calculate A3 and A4 for 0. 1 0 A =| 0 0 1 1 -4 4

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

The Cayley-Hamilton Theorem provides a method for calculating powers of a matrix. For example, if $ A $ is a $ 3 \times 3 $ matrix with the characteristic equation

\[ c_0 + c_1 \lambda + c_2 \lambda^2 + \lambda^3 = 0 \]

then $ c_0I + c_1A + c_2A^2 + A^3 = 0 $, so

\[ A^3 = -c_2A^2 - c_1A - c_0I \]

Multiplying through by $ A $ yields

\[ A^4 = -c_0A - c_1A^2 - c_2A^3 \]

which expresses $ A^4 $ in terms of $ A^3, A^2 $, and $ A $. Use this procedure to calculate $ A^3 $ and $ A^4 $ for

\[ A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -4 & 4 \end{bmatrix} \]

$ A^3 $ is

\[
\begin{bmatrix}
\boxed{} & \boxed{} & \boxed{} \\
\boxed{} & \boxed{} & \boxed{} \\
\boxed{} & \boxed{} & \boxed{}
\end{bmatrix}
\]

$ A^4 $ is

\[
\begin{bmatrix}
\boxed{} & \boxed{} & \boxed{} \\
\boxed{} & \boxed{} & \boxed{} \\
\boxed{} & \boxed{} & \boxed{}
\end{bmatrix}
\]

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