6.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX 147
and the row reduced echelon form is 1 0 −1 0
0 1 0 0
0 0 0 0
Therefore, the eigenvectors are of the form
z
1
0
1
where z ̸= 0.
Next find the eigenvectors for λ = 2. The augmented matrix for the system of equationsneeded to find these eigenvectors is 0 −2 2 0
−1 −1 1 0
1 −1 1 0
and the row reduced echelon form is 1 0 0 0
0 1 −1 0
0 0 0 0
and so the eigenvectors are of the form
z
0
1
1
where z ̸= 0.
Finally find the eigenvectors for λ = 4. The augmented matrix for the system of equationsneeded to find these eigenvectors is 2 −2 2 0
−1 1 1 0
1 −1 3 0
and the row reduced echelon form is 1 −1 0 0
0 0 1 0
0 0 0 0
.
Therefore, the eigenvectors are of the form
y
1
1
0
where y ̸= 0.