(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:

(l

^{2}-3l +2) – 6 = 0 or l^{2}-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:

**k1**]

[2 ...2] [

**k2**]

3

**k1**+ 3**k2**= 0, and 2**k1**+ 2**k2**= 0So:

**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: -2

**k1**+ 3

**k2**= 0 and 2

**k1**- 3

**k2**= 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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