Question

\[ \left( \begin{array}\ 1 & 0 &0 \\ 1 &0& 1\\ 0& 1 & 0 \end{array}\right)\] =A by cayley hamilton transformation method prove that \[A^n=A^n-2 +A^2-1\] and find \[A^50\]

Answer 1

Sorry my question is not appearing...

Answer 2

In a 3*3 matrix with a11=1,a12=0,a13=0,a21=1,a22=0,a23=1,a31=0,a32=1,a33=0...prove by cayley hamilton theorem that A^n= A^n-2+A^2-1.....also find A^50

Answer 3

@Jemurray3 @

Answer 4

The Cayley-Hamilton theorem says that matrices obey their own characteristic equations. So,
\[det (A-\lambda I) = det\left(\begin{matrix}1-\lambda & 0 & 0 \\ 1 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{matrix}\right) = \lambda^2(1-\lambda) -(1-\lambda) \]
\[ = (\lambda^2 -1)(1-\lambda) = -\lambda^3+ \lambda^2 + \lambda - 1 = 0 \]
implies that
\[ -A^3 + A^2 + A - I = 0 \implies A^3 = A^2+ A - I \]

We can use a proof by induction. Assume
\[A^n = A^{n-2} + A^2 - I \]
Then
\[A^{n+1} = A\cdot A^n = A^{n-1} + A^3- A = A^{n-1} + A^2 + A - I - A\]
\[ = A^{(n+1)-2} + A^2 - I \]
Since we've already proved this for the case n = 2, it must hold for all n >= 2.

Answer 5

Furthermore, by the above theorem,
\[A^{50} = A^{48} + A^2 - I = A^{46} + 2 ( A^2- I ) = A^{44} + 3(A^2- I) \]
etc etc etc. Which yields the general formula (assuming that k is even)
\[ A^k = A^2 + \frac{k-2}{2} \left( A^2 - I \right) \]
So
\[ A^{50} = A^2 + 24( A^2 - I ) = 25 A^2 - 24I\]
A^2 is fairly easy to calculate:
\[A^2 = \left(\begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}\right)\]
So finally,
\[A^{50} = \left( \begin{matrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 &1\end{matrix} \right) \]

Answer 6

but how did \[A^{48}+A^2-I\] turn into \[A^{46}+2(A^2-I)\]

Answer 7

Because via the proof above,
\[ A^k = A^{k-2} + A^2 - I \]

Answer 8

thank you:)