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Verify that A=$\begin{bmatrix}1&-1\\0&2\end{bmatrix}$ satisfies $A^2-3A+2I=0$.

Use this fact to show that $A^{-1}= \frac{1}{2}(3I-A)$

I know it does satisfies the first equation/already proved it.

Spectre
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  • Visit this page to learn how to properly type mathematics on this site. Typing a matrix for instance can be accomplished as $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ to produce $\begin{bmatrix}a&b\c&d\end{bmatrix}$ – JMoravitz Oct 22 '20 at 12:52
  • Welcome to MathSE. Please, can you use MathJax for to write your problem? I don't understand your notation. –  Oct 22 '20 at 12:52
  • Now, as for the content of your question, What is $A\cdot \frac{1}{2}(3I-A)$? Use what you verified already about $A^2-3A+2I=0$ for our $A$. Do you find that $A\cdot \frac{1}{2}(3I-A)$ is equal to $I$? What does that mean? Recall your basic definitions of inverse matrices. – JMoravitz Oct 22 '20 at 12:55

2 Answers2

1

By the given equation, we have $A^2-3A=-2I$, thus $$\frac12(3A-A^2)=I$$

Berci
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An $n\times n$ square matrix $A$ is said to be invertible iff there exists an $n\times n$ matrix $B$ such that $AB = I_{n\times n}$.

If such a matrix $B$ exists, it is known that it will be unique and we can give it the name "The inverse of $A$" and a special symbol, $A^{-1}$, to further emphasize the relationship between it and the original matrix $A$. It is also known that for square matrices that $AB=I$ will imply that $BA=I$. (Warning: the same is not true for non-square matrices)

For your problem, once having verified that our matrix $A$ satisfies the matrix equation mentioned, all that remains to prove that $\frac{1}{2}(3I-A)$ is in fact the inverse of $A$ is to show that $A\cdot \frac{1}{2}(3I-A) = I$.

By basic algebraic manipulation we have that $$A\cdot \frac{1}{2}(3I-A) = \frac{1}{2}(3A-A^2)$$

From here, we can use the fact that "adding zero" doesn't change anything and that from the earlier part of the problem we learned that $0 = A^2-3A+2I$, so this continues:

$$\frac{1}{2}(3A-A^2) = \frac{1}{2}(3A-A^2+0) = \frac{1}{2}(3A-A^2+A^2-3A+2I) = \frac{1}{2}(2I)=I$$

We have thus shown that $A\cdot \frac{1}{2}(3I-A)=I$, thereby proving that $\frac{1}{2}(3I-A)=A^{-1}$

JMoravitz
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