????

How did mysteriously change into ?

Is one of those a typo?

Okay, the first has eigenvalues 1 and -6 as you say while the second has eigenvalues 6 and -4 so I assume the second was probably accidently copied from a different problem.

Notice an important difference between this and a similar problem that was posted (perhaps by you?) earlier on this board. Before the problem asked for "invariant lines" which meant, of course, lines that are mapped into themselves. Those turn out to be the lines in the "characteristic directions" or in the directions given by the eigenvectors. But saying that alineis "invariant" doesn't mean that points on that line are mapped into themselves. It means points on that line are mapped into other points on that same line.

Here, you are asked for a line on which "every point is mapped into itself". That is, you are asked for "invariant points" which happen to lie on the line corresponding to eigenvalue1.

For example, the point (1, 2) is on the line y= 2x. If you apply the matrix to that, , a different point because each component is multiplied by -6, but on the same line. If we take the point (3, 5), on the line 3y= 5x, and do the same, , exactly the same point: sure, it has been "multiplied" by its eigenvalue,1!

Any eigenvector gives an "invariant line" but only eigenvectors corresponding to eigenvalue 1 give "invariant points".