Eigenvectors and Inverse Laplace Transforms

Let's work through this university-level mathematics problem. We have two main parts: finding eigenvalues and eigenvectors for a matrix, and then computing some inverse Laplace transforms.

First, we need to show that the determinant of A minus lambda I is zero to find the eigenvalues. This is the characteristic equation.

Let's expand this determinant. Expanding along the third column is convenient because of the zero. We have one times the minor, minus zero, plus negative one minus lambda times its minor.

Wait, let's actually expand along the second row to keep things consistent purely algebraically. Let's rewrite the full expansion carefully.

Let's simplify. The first term is one plus lambda, squared. The second term gives plus twelve times one plus lambda. And the last term simplifies to negative twelve plus, well, let's group it.

Actually, notice that row 1 plus row 3 in the original matrix is zero, which means the determinant is zero so lambda equals zero is a root. Let's collect terms for the full polynomial.

We can factor out negative lambda. This leaves us with a quadratic: lambda squared plus lambda minus twelve.

Factoring the quadratic gives us lambda plus four and lambda minus three. So, our eigenvalues are zero, three, and negative four.

Now we find the eigenvectors for each lambda. Let's start with lambda equals zero. We solve A times v equals zero.

From the second row, we see that six x minus y is zero, so y equals six x. From the first row, x plus two y plus z is zero.

Substituting y equals six x into the second equation: x plus twelve x plus z is zero. That means z equals negative thirteen x.

Picking x equal to one, we get our first eigenvector.

Next, for lambda equals three. We subtract 3 from the diagonal elements.

The second row gives six x minus four y equals zero. Dividing by two, we get three x equals two y. Let's choose x equals two, so y equals three.

Now use the first row to find z. Negative two times two, plus two times three, plus z equals zero. That's minus four plus six plus z equals zero.

So our second eigenvector is 2, 3, negative 2.

Finally, for lambda equals negative four. Adding 4 to the diagonal gives us this matrix.