A=⎣⎡21311−11−11⎦⎤
Characteristic polynomial of A,
det(A−λI)=0
⇒det⎝⎛⎣⎡21311−11−11⎦⎤−λ⎣⎡100010001⎦⎤⎠⎞=0
⇒det⎝⎛⎣⎡2−λ1311−λ−11−11−λ⎦⎤⎠⎞=0
⇒(2−λ)det([1−λ−1−11−λ])
−det([13−11−λ])
+det([131−λ−1])=0
⇒(2−λ)(λ2−2λ)−(−λ+4)+(3λ−4)=0
⇒λ3−4λ2+8=0
∴ Characteristic polynomial of A :- λ3−4λ2+8=0
Now,
A2=A×A=⎣⎡21311−11−11⎦⎤×⎣⎡21311−11−11⎦⎤
=⎣⎡8082312−15⎦⎤
and,
A3=A2×A=⎣⎡8082312−15⎦⎤×⎣⎡21311−11−11⎦⎤
=⎣⎡240328448−412⎦⎤
Now, by Cayley-Hamilton theorem, putting λ=A ,
A3−4A2+8I=03×3
⇒⎣⎡240328448−412⎦⎤−4⎣⎡8082312−15⎦⎤
+8⎣⎡100010001⎦⎤=03×3
⇒⎣⎡000000000⎦⎤=03×3
∴ Cayley-Hamilton Theorem Verified.
Now, by Cayley-Hamilton theorem,
A3−4A2+8I=03×3
⇒A(A3−4A2+8I)=A×03×3
[ Multiplying both side by matrix A ]
⇒A4−4A3+8A=03×3
⇒A4=4A3−8A
=4⎣⎡240328448−412⎦⎤−8⎣⎡21311−11−11⎦⎤ =⎣⎡80−81042482424−840⎦⎤
∴A4=⎣⎡80−81042482424−840⎦⎤
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