∣ A − λ I = 0 e q u a t i o n 1 − λ 3 + λ 2 + 18 λ + 30 = 0 − λ 3 + λ 2 + 18 λ + 30 I = 0 C a l c u l a t e A 2 = [ 14 3 14 12 9 6 8 6 14 ] C a l c u l a t e A 3 = [ 62 39 68 48 21 78 62 24 62 ] P u t a l l v a l u e s i n e q u a t i o n 1 = [ − 62 − 39 − 68 − 48 − 21 − 78 − 62 − 24 − 62 ] + [ 14 3 14 12 9 6 8 6 14 ] + 18 [ 1 2 3 2 − 1 4 3 1 1 ] + 30 [ 1 0 0 0 1 0 0 0 1 ] = [ 0 0 0 0 0 0 0 0 0 ] L . H . s = R . H . s S o t h i s m a t r i x m e e t s C a l e y H a m i l t o n T h e o r e m \mid A- \lambda I=0\\ equation 1
- \lambda^3+\lambda^2+18\lambda+30=0\\
- \lambda^3+\lambda^2+18\lambda+30I=0\\
Calculate \, A^2=\begin{bmatrix}
14 & 3&14\\
12 & 9&6\\
8&6&14\\
\end{bmatrix}\\
Calculate \, A^3=
\begin{bmatrix}
62& 39&68\\
48 & 21&78\\
62&24&62\\
\end{bmatrix}
\\Put \, all \, values \, in \, equation \, 1\\\\=
\begin{bmatrix}
-62& -39&-68\\
-48 & -21&-78\\
-62&-24&-62
\end{bmatrix}+
\begin{bmatrix}
14& 3&14\\
12 & 9&6\\
8&6&14\\
\end{bmatrix}+18
\begin{bmatrix}
1& 2&3 \\
2& -1&4\\
3&1&1\\
\end{bmatrix}+30
\begin{bmatrix}
1& 0&0 \\
0& 1&0\\
0&0&1\\
\end{bmatrix}\\
=
\begin{bmatrix}
0&0&0\\
0&0&0\\
0&0&0
\end{bmatrix}\\
L.H.s=R.H.s\\
So\, this\, matrix\, meets \,Caley \, Hamilton\, Theorem ∣ A − λ I = 0 e q u a t i o n 1 − λ 3 + λ 2 + 18 λ + 30 = 0 − λ 3 + λ 2 + 18 λ + 30 I = 0 C a l c u l a t e A 2 = ⎣ ⎡ 14 12 8 3 9 6 14 6 14 ⎦ ⎤ C a l c u l a t e A 3 = ⎣ ⎡ 62 48 62 39 21 24 68 78 62 ⎦ ⎤ P u t a ll v a l u es in e q u a t i o n 1 = ⎣ ⎡ − 62 − 48 − 62 − 39 − 21 − 24 − 68 − 78 − 62 ⎦ ⎤ + ⎣ ⎡ 14 12 8 3 9 6 14 6 14 ⎦ ⎤ + 18 ⎣ ⎡ 1 2 3 2 − 1 1 3 4 1 ⎦ ⎤ + 30 ⎣ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎤ = ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ L . H . s = R . H . s S o t hi s ma t r i x m ee t s C a l ey H ami lt o n T h eore m
#339914 on Dec 2023
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