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

Dear Student,
Please find below the solution to the asked query:

We have:A=2-334A2=2-3342-334=4-9-6-126+12-9+16=-5-18187L.H.S.=x2-6x+17I, where I is indentity matrix.Put x=AL.H.S.=A2-6A+17I=-5-18187-62-334+171001=-5-18187+-1218-18-24+170017=-5-12+17-18+18+018-18+07-24+17=0000=R.H.S.Hence A satisfies given equation.HenceA2-6A+17I=0AA-6I+17A-1=0As A0, hence we get:A-6I+17A-1=017A-1=6I-A17A-1=61001-2-334=6006-2-33417A-1=43-32A-1=417317-317217

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards

  • 10
@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 ]
  • -3
What are you looking for?