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Is the following statement True/false ?

Let $M = \begin{bmatrix} 1 &-1& 1\\2 &1 &4 \\-2 &1 &-4 \end{bmatrix}$ then $M^{-1} = \frac{1}{4} (M +3I)$

My attempt : I think this statement is true because $1$ is an eigenvalue of $M$. Also, the minimal polynomial of $M$ is $(x-1)(x+4)$.

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    $\begingroup$ the minimal polynomial: $(x-1)^2(x+4)$ $\endgroup$
    – Lozenges
    Commented Mar 6, 2023 at 7:13
  • $\begingroup$ yes, it is the minimal polynomial as well $\endgroup$
    – Lozenges
    Commented Mar 6, 2023 at 7:17
  • $\begingroup$ Have you tried to verify $MM^{-1}=M^{-1}M=I$? $\endgroup$
    – CroCo
    Commented Mar 6, 2023 at 9:32
  • $\begingroup$ okay @CroCo No, I didn't use this logic $\endgroup$
    – jasmine
    Commented Mar 6, 2023 at 16:49

2 Answers 2

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The polynomial $$P(\lambda)=(\lambda-1)(\lambda+4)$$ is not the minimal polynomial of $M$.

A quick calculation shows that

$$(M-I_3)(M+4I_3)\neq 0$$ from which we find $$M^2+3M-4I_{3}\neq 0\implies M^{-1}\neq \frac{1}{4}(M+3I_3)$$

Remark

Thanks to CroCo and Ted Shifrin's kind comments, I based my previous answer on OP's assumptions.

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    $\begingroup$ Isn't the characteristic polynomial of degree $3$ for this matrix? $\endgroup$ Commented Mar 6, 2023 at 6:11
  • $\begingroup$ @UmeshShankar I meant minimal. Thanks ! $\endgroup$
    – Hamdiken
    Commented Mar 6, 2023 at 6:13
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    $\begingroup$ I've tried yours, I got $M^{-1}M\neq I$. $\endgroup$
    – CroCo
    Commented Mar 6, 2023 at 10:07
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    $\begingroup$ This is incorrect. Try calculating $M^2+3M-4I$. In fact, $1$ is an eigenvalue with geometric multiplicity $1$, not $2$. $\endgroup$ Commented Mar 6, 2023 at 16:02
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    $\begingroup$ Basing an answer on the OP's assumptions does not often lead to a good answer. $\endgroup$ Commented Mar 6, 2023 at 18:00
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The characteristic equation of $M$ is $$ p(\lambda) = \lambda^3 + 2\lambda^2-7\lambda + 4 $$ Using Cayley–Hamilton theorem, we get $$ \begin{align} M^3 + 2M^2-7M + 4I = 0 \implies M^{-1}=(-1/4)\Big( M^2 + 2M -7I\Big). \end{align} $$ Indeed, $MM^{-1}=M^{-1}M=I$.

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  • $\begingroup$ The downvotes are not clear to me. $\endgroup$
    – CroCo
    Commented Mar 6, 2023 at 11:16
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    $\begingroup$ This is actually correct. The claim below about the minimal polynomial of $M$ is, in fact, incorrect. $\endgroup$ Commented Mar 6, 2023 at 16:03

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