# Is $M^{-1} = \frac{1}{4} (M +3I)$? yes/ No

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 $$(x-1)(x+4)$$.

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

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.

• Isn't the characteristic polynomial of degree $3$ for this matrix? Commented Mar 6, 2023 at 6:11
• @UmeshShankar I meant minimal. Thanks ! Commented Mar 6, 2023 at 6:13
• I've tried yours, I got $M^{-1}M\neq I$. Commented Mar 6, 2023 at 10:07
• This is incorrect. Try calculating $M^2+3M-4I$. In fact, $1$ is an eigenvalue with geometric multiplicity $1$, not $2$. Commented Mar 6, 2023 at 16:02
• Basing an answer on the OP's assumptions does not often lead to a good answer. Commented Mar 6, 2023 at 18:00

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, $$MM^{-1}=M^{-1}M=I$$.

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