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$$A^2=A-2I$$ Does $A^{-1}$ exist if so calculate it in terms of $A$.

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    What does $-2$ mean? – 2017-01-11
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    Hint: rewrite your equation to get $A(I-A)=2I$. – 2017-01-11
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    @Batman presumably, $-2$ means $-2I$. – 2017-01-11
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    @MichaelBurr I'm inclined to think Batman wanted to hear that from the OP. After all, this post has many of the hallmarks of a question soon to be put on hold for lack of context. Kriss, care to comment? Where did you run into this question? What do $-2$ and $A^-$ mean here? they don't really make sense as matrices, so for you to understand any answer you should be able to clarify. – 2017-01-11
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    @MichaelBurr - yep, what Jykri said. Just trying to get a decently written question. – 2017-01-11
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    @Batman srry for not looking i edited it and the matrix is 3x3 – 2017-01-13

1 Answers 1

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Supposing that $A$ is a $n\times n$ matrix and $-2$ means $-2I$:

$$A^2=A-2I \rightarrow A(A-I)=-2I \rightarrow \det(A)\det(A-I)=(-2)^n$$

So $\det(A)\ne 0$.

For finding $A^{-1}$ as a function of $A$ multiply both sides by $A^{-1}$ and get:

$$A=I-2A^{-1} \rightarrow A^{-1}=\frac{1}{2}(I-A)$$

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    @Nick Liu Why did you delete your answer: it was correct ! – 2017-01-11