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 December 18th, 2017, 09:50 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 Finding matrix A inverse... If A is an n x n matrix and it satisfies the equation: $\displaystyle A^3 - 4A^2 + 3A - 5I_n = 0$ then A is non-singular and its inverse is ??????
 December 18th, 2017, 10:52 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics Its inverse is $\frac{1}{5}\left( A^2 -4A + 3 \right)$. Do you see why? Thanks from Country Boy and zollen
December 18th, 2017, 10:57 AM   #3
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Quote:
 Originally Posted by SDK Its inverse is $\frac{1}{5}\left( A^2 -4A + 3 \right)$. Do you see why?
I like this problem. Trivially easy once you step outside the box.

December 18th, 2017, 10:57 AM   #4
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Quote:
 Originally Posted by SDK Its inverse is $\frac{1}{5}\left( A^2 -4A + 3 \right)$. Do you see why?
$\displaystyle 5I = A * (A^2 - 4A + 3)$
$\displaystyle I = \frac{1}{5} ( A * ( A^2 - 4A + 3))$
$\displaystyle A^{-1}I = \frac{1}{5} ( A^2 - 4A + 3)$

December 18th, 2017, 11:21 AM   #5
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Quote:
 Originally Posted by zollen $\displaystyle 5I = A * (A^2 - 4A + 3)$ $\displaystyle I = \frac{1}{5} ( A * ( A^2 - 4A + 3))$ $\displaystyle A^{-1}I = \frac{1}{5} ( A^2 - 4A + 3)$
Very good!
Equivalently, $\displaystyle A^3- 4A^2+ 3A- 5I= 0$ so
$\displaystyle A^3- 4A^2+ 3A= 5I$
$\displaystyle A(A^2- 4A+ 2)= (A^2- 4A+ 3)A= 5I$

$\displaystyle A((1/5)(A^2- 4A+ 3))= ((1/5)(A^2- 4A+ 3)A= I$.

That is sufficient to show that $\displaystyle (1/5)(A^2- 4A+ 3)$ is the inverse of A.

Last edited by skipjack; December 18th, 2017 at 05:16 PM.

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