Verify Cayley hamilton theorem for a matrix A=(1 2 3 3 1 -1 0 2 -1)
Answers
Answer:
Step-by-step explanatiA=\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}A=
⎣
⎢
⎡
2
1
3
1
1
−1
1
−1
1
⎦
⎥
⎤
Characteristic polynomial of A,
det(A-\lambda I)=0det(A−λI)=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
⎝
⎛
⎣
⎢
⎡
2
1
3
1
1
−1
1
−1
1
⎦
⎥
⎤
−λ
⎣
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎤
⎠
⎞
=0
\Rightarrow det \left(\begin{bmatrix}2-\lambda&1&1\\1&1-\lambda&-1\\3&-1&1-\lambda\end{bmatrix}\right)=0⇒det
⎝
⎛
⎣
⎢
⎡
2−λ
1
3
1
1−λ
−1
1
−1
1−λ
⎦
⎥
⎤
⎠
⎞
=0
\Rightarrow (2-\lambda)det\left(\begin{bmatrix}1-\lambda&-1\\-1&1-\lambda\end{bmatrix}\right)⇒(2−λ)det([
1−λ
−1
−1
1−λ
])
-det\left(\begin{bmatrix}1&-1\\3&1-\lambda\end{bmatrix}\right)−det([
1
3
−1
1−λ
])
+det\left(\begin{bmatrix}1&1-\lambda\\3&-1\end{bmatrix}\right)=0+det([
1
3
1−λ
−1
])=0
\Rightarrow \left(2-\lambda\right)\left(\lambda^2-2\lambda\right)-\left(-\lambda+4\right)+ \left(3\lambda-4\right)=0⇒(2−λ)(λ
2
−2λ)−(−λ+4)+(3λ−4)=0
\Rightarrow \lambda^3-4\lambda^2+8=0⇒λ
3
−4λ
2
+8=0
\therefore∴ Characteristic polynomial of A :- \lambda^3-4\lambda^2+8=0λ
3
−4λ
2
+8=0
Now,
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}A
2
=A×A=
⎣
⎢
⎡
2
1
3
1
1
−1
1
−1
1
⎦
⎥
⎤
×
⎣
⎢
⎡
2
1
3
1
1
−1
1
−1
1
⎦
⎥
⎤
=\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}=
⎣
⎢
⎡
8
0
8
2
3
1
2
−1
5
⎦
⎥
⎤
and,
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}A
3
=A
2
×A=
⎣
⎢
⎡
8
0
8
2
3
1
2
−1
5
⎦
⎥
⎤
×
⎣
⎢
⎡
2
1
3
1
1
−1
1
−1
1
⎦
⎥
⎤
=\begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}=
⎣
⎢
⎡
24
0
32
8
4
4
8
−4
12
⎦
⎥
⎤
Now, by Cayley-Hamilton theorem, putting \lambda=Aλ=A ,
A^3-4A^2+8I=0_{3\times3}A
3
−4A
2
+8I=0
3×3
\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}⇒
⎣
⎢
⎡
24
0
32
8
4
4
8
−4
12
⎦
⎥
⎤
−4
⎣
⎢
⎡
8
0
8
2
3
1
2
−1
5
⎦
⎥
⎤
+8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=0_{3\times3}+8
⎣
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎤
=0
3×3
\Rightarrow\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0_{3\times3}⇒
⎣
⎢
⎡
0
0
0
0
0
0
0
0
0
⎦
⎥
⎤
=0
3×3
\therefore∴ Cayley-Hamilton Theorem Verified.
Now, by Cayley-Hamilton theorem,
A^3-4A^2+8I=0_{3\times3}A
3
−4A
2
+8I=0
3×3
\Rightarrow A(A^3-4A^2+8I)=A\times0_{3\times3}⇒A(A
3
−4A
2
+8I)=A×0
3×3
[ Multiplying both side by matrix AA ]
\Rightarrow A^4-4A^3+8A=0_{3\times3}⇒A
4
−4A
3
+8A=0
3×3
\Rightarrow A^4 = 4A^3-8A⇒A
4
=4A
3
−8A
=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}=4
⎣
⎢
⎡
24
0
32
8
4
4
8
−4
12
⎦
⎥
⎤
−8
⎣
⎢
⎡
2
1
3
1
1
−1
1
−1
1
⎦
⎥
⎤
=\begin{bmatrix}80&24&24\\-8&8&-8\\104&24&40\end{bmatrix}=
⎣
⎢
⎡
80
−8
104
24
8
24
24
−8
40
⎦
⎥
⎤
\therefore A^4=\begin{bmatrix}80&24&24\\-8&8&-8\\104&24&40\end{bmatrix}∴A
4
=
⎣
⎢
⎡
80
−8
104
24
8
24
24
−8
40
⎦
⎥
⎤
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