Problem14:Consider the matrices, 
> ;
[2 1 2] [ ] A := [2 2 2] [ ] [3 1 1] [x1] [ ] x := [x2] [ ] [x3]
Show that the equation Ax=x can be rewritten as (AI)x=0 and use
this result to solve Ax=x for x. Also solve Ax=4x.

It is obvious that we can write Ax=x as (AI)x = 0. Just substract x from both sides and use the property that Ix=x and the distributivity of matrix multiplication over matrix addition. To find x we can bring the augmented matrix of the homogeneous system (AI)x=0 into rref (reduced row echelon form). Here is one way to do it with maple, 
> A := matrix(3,3,[2,1,2,2,2,2,3,1,1]); I3 := diag(1,1,1);
[2 1 2] [ ] A := [2 2 2] [ ] [3 1 1] [1 0 0] [ ] I3 := [0 1 0] [ ] [0 0 1]> rref( augment(AI3, matrix(3,1,[0,0,0])) );
[1 0 0 0] [ ] [0 1 0 0] [ ] [0 0 1 0]
this shows that the only solution for x is the trivial solution x=0. i.e. x1=x2=x3=0. Not very interesting but look at the next case. Let us try to find x such that Ax=4x. Now we can find infinitely many solutions... As before we bring into rref the augmented matrix for the homogeneous system (A4I)x=0. 
> rref( augment( A  4*I3, matrix(3,1,[0,0,0]) ) );
[1 0 1 0] [ ] [0 1 0 0] [ ] [0 0 0 0]
this system says that any x such that x1=x3 and that x2=0 will do.
In other words t*(1,0,1) for any scalar t will be a solution.
These kinds of problems (i.e. solving Ax=kx for some scalar k) are very important in linear algebra. Non trivial x's (i.e. different from 0) of this kind are known as eigen vectors for the matrix A associated to the eigen value k. We'll study eigen values and eigen vectors later on in this course when we talk about diagonalization of matrices. 