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If a matrix $A$ satisfies the equation $A^3-6 A^2+11 A-6 I=0$, then $A^{-1}$ can be
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Verified Answer
The correct answer is:
$\frac{1}{3} /$
$$
\begin{aligned}
& A^3-6 A^2+11 A-6 I=0 \\
& A^3-A^2-5 A^2+5 A+6 A-6 I=0 \\
& \Rightarrow A^2[A-I]-5 A[A-I]+6[A-I]=0 \\
& \Rightarrow {[A-I]\left[A^2-5 A+6\right]=0 } \\
& \Rightarrow(A-I)(A-2 I)(A-3 I)=0
\end{aligned}
$$
$\therefore A-3 I$ can be a zero matrix.
$$
\begin{aligned}
\Rightarrow & & A & =3 I \\
\Rightarrow & & A A^{-1} & =3 A^{-1} \\
\Rightarrow & & I & =3 A^{-1} \\
\Rightarrow & & A A^{-1} & =\frac{1}{3} I
\end{aligned}
$$
\begin{aligned}
& A^3-6 A^2+11 A-6 I=0 \\
& A^3-A^2-5 A^2+5 A+6 A-6 I=0 \\
& \Rightarrow A^2[A-I]-5 A[A-I]+6[A-I]=0 \\
& \Rightarrow {[A-I]\left[A^2-5 A+6\right]=0 } \\
& \Rightarrow(A-I)(A-2 I)(A-3 I)=0
\end{aligned}
$$
$\therefore A-3 I$ can be a zero matrix.
$$
\begin{aligned}
\Rightarrow & & A & =3 I \\
\Rightarrow & & A A^{-1} & =3 A^{-1} \\
\Rightarrow & & I & =3 A^{-1} \\
\Rightarrow & & A A^{-1} & =\frac{1}{3} I
\end{aligned}
$$
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