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December 18th, 2017, 08:50 AM   #1
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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 ??????
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December 18th, 2017, 09:52 AM   #2
SDK
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Its inverse is $\frac{1}{5}\left( A^2 -4A + 3 \right)$. Do you see why?
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December 18th, 2017, 09:57 AM   #3
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Quote:
Originally Posted by SDK View Post
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.
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December 18th, 2017, 09:57 AM   #4
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Quote:
Originally Posted by SDK View Post
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)
$
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December 18th, 2017, 10:21 AM   #5
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Quote:
Originally Posted by zollen View Post
$\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.
Thanks from zollen

Last edited by skipjack; December 18th, 2017 at 04:16 PM.
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