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matrix problem... Pls help...?
A certain 2 x 2 matrix P has 1 as one of its eigenvalues.
Determine the corresponding eigenvalue of the matrix [ (p square + 3I] inverse. Give your ans to 3 decimal place.
Thanks.
2 Answers
- kbLv 71 decade agoFavorite Answer
By definition of an eigenvalue, Pv = 1v for some nonzero (eigen)vector v.
Thus, P^2 v = P (Pv) = P (1v) = 1 * Pv = 1^2 v = v.
So, (P^2 + 3I)v = P^2v + 3Iv = v + 3v = 4v.
Assuming the invertibility of (P^2 + 3I), multiplying both sides on the left by (P^2 + 3I)^(-1) yields
v = (P^2 + 3I)^(-1) (4v)
==> v = 4 (P^2 + 3I)^(-1) (v)
==> (1/4) v = (P^2 + 3I)^(-1) (v)
Thus, 1/4 = 0.25 is an eigenvalue (P^2 + 3I)^(-1) (with the same eigenvector v).
I hope that helps!
- nleLv 71 decade ago
I use the matrix
[3/2 1 ]
[1 3 ]
which has the eigenvalue 7/2 and 1
and ( P^2 + 3I) ^(-1)
=[ 13/61 -9/122 ]
[ -9/122 25/244 ]
which has eigenvalues 1/4 and 4/61
That proved kb is correct.
