(2 .....3)
(2......1)
Then, it must be true from the properties of determinants that: (A - lI) D =
[(2 - l)......3] [k1]
[2 .......(1 -l)] [k2]
Note how we allow l ('lambda') to be subtracted from the first element in the upper left, and from the last element in the lower right). Cross-multiplying and using matrix properties we obtain the characteristic equation:
(l2 -3l +2) – 6 = 0 or l2 -3l -4 = 0
So that: (l - 4) (l + 1) = 0, where l1 = -1 and l2 = 4
We need to find a vector that solves the equation: (A - lI) D = 0
In the first instance, we substitute the first eigenvalue, l1= -1, into the matrix, whence:
[2 ...2] [k2]
3k1 + 3 k2 = 0, and 2k1 + 2 k2 = 0
So: k1 =1 and k2 = -1
The first eigenvector is then: K1 =
[1]
So: X1 = K1 exp (-t)
The second eigenvalue was l2 = 4 so we repeat the process again to obtain the equation to be solved:
(A - l2 I)D = (A - (4)I) D = (A + I)D = 0 =
[-2.....3] [k1]
[2 ...-.3] [k2]
Or: -2k1 + 3k2 = 0 and 2 k1 - 3k2 = 0
So that: k1 = 3 and k2 = 2
The second eigenvector is then: K2 =
[3]
[2]
And another linearly independent solution is: X2 = K2 exp (4t)
Because of the principle of linear superposition, one can also add the two solutions, to obtain:
X = X1 + X2 = K1 exp (-t) + K2 exp (4t)
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