Prove a matrix is invertible

Question

The $2 \times 2$ matrix ${A}$ satisfies $A^2 - 4 {A} - 7{I} = {0},$ where ${I}$ is the $2 \times 2$ identity matrix. Prove that ${A}$ is invertible.

What is the best way to do this?

use the definition of invertible matrices – Oussama Boussif Aug 10 '15 at 22:30 — Oussama Boussif, Aug 10 '15 at 22:30

score 8 · Answer 1 · edited Aug 10 '15 at 22:47

8

$$ A^2-4A-7I = 0\\ A(A-4I) = 7I\\ A\left(\frac{A-4I}{7}\right) = I $$

A matrix $A$ is invertible if and only if there exist ${A}^{-1}$ such that:

$$ A{A}^{-1}= I $$

So from our previous answer we conclude that:

$$ {A}^{-1} = \frac{A-4I}{7} $$

So ${A}^{-1}$ exists, hence $A$ is invertible.

Note:

if you had the value of $A$ you would only calculate its determinant and check if it is non zero.

$\det(A)\quad \neq \quad 0\Longleftrightarrow A\quad \textrm{is}\quad \textrm{invertible}$

edited Aug 10 '15 at 22:47

johannesvalks

answered Aug 10 '15 at 22:34

Oussama Boussif

1

Nice answer. Saves me the work to answer this OP. +1 I made a small edit... – johannesvalks Aug 10 '15 at 22:49
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Ah thanks, always glad to help ^^ – Oussama Boussif Aug 10 '15 at 22:49
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Tip for $\LaTeX$, use $\backslash\textrm{det}$, and $\backslash\textrm{left}($ $\backslash\textrm{right})$... Good Luck! – johannesvalks Aug 10 '15 at 22:51
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Oh thanks that will help write more organised answers ;) – Oussama Boussif Aug 10 '15 at 22:54

score 2 · Answer 2 · answered Aug 10 '15 at 22:33

2

$${1\over7}(A-4I)A={1\over 7}(A^2-4A)={1\over 7}{7I}=I\\ A{1\over7}(A-4I)={1\over 7}(A^2-4A)={1\over 7}{7I}=I$$

answered Aug 10 '15 at 22:33

score 1 · Answer 3 · answered Aug 10 '15 at 23:25

Given that $A$ satisfies $A^2−4A−7I=0$ (use Cayley- Hamilton theorem). Let $\gamma_1$ and $\gamma_2$ be eigen value of $A$.

$\gamma_1+\gamma_2=4$.

$\gamma_1\gamma_2=-7$. Therefore $\gamma_1$ and $\gamma_2$ are non zero. Hence $A$ is invertible matrix

3 Answers3