# Show that  A = [  2    -3          3     4   ] satisfies the equation x^2 - 6x + 17 = 0. Hence find A^-1.

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• 7
@Amir

In matrix algebra the equation mentioned in the question has the form -
x^2 - 6x + 17I = 0

where I is the 2x2 identity matrix in this case.

substituting x with the Matrix A we have

x^2 = A A = [-5 -18
18 7]

6x = 6 A = [12 -18
18 24]

17 = 17I = [17 0
[0 17]

with the above matrices u can verify that matrix A satisfies the equation given.

Now to find A^-1 , we have

--- AA - 6A + 17I = 0
--- A (A - 6 + 17 A^-1) = 0 ,
using AA^-1 = I
--- now since A is not equal to 0
we have
-- 17A^-1 = 6 I - A

therefore
A^-1 = (1/17)(6I - A)

which on substituting the value of A and identity matrix I we have

A^-1 = [4/17 3/17
-3/17 2/17 ]
• -4
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