Attachment 38634
ok im doing #1
ok I go a reduction of
$$\frac{1}{3}
\begin{bmatrix}
1&1/3&0\\
0&1&0\\
0&0&1
\end{bmatrix}$$
but what next?
the bk answer is
$-\lambda^3 +18\lambda^2
-95-\lambda+150$
Attachment 38634
ok im doing #1
ok I go a reduction of
$$\frac{1}{3}
\begin{bmatrix}
1&1/3&0\\
0&1&0\\
0&0&1
\end{bmatrix}$$
but what next?
the bk answer is
$-\lambda^3 +18\lambda^2
-95-\lambda+150$
You say, in your post, that you are doing #1 but there is no "#1" in the attachment. Instead, #13 is highlighted so I assume that is what you mean.
Your problem is that while "reducing" a matrix to upper triangular form is useful for finding an inverse matrix or solving matrix equations, it has nothing to do with finding eigenvalues. The eigenvalues of the reduced matrix are not those of the original matrix.
Instead we need to solve the determinant equation $\displaystyle \left|\begin{array}{ccc}6-\lambda & -2 & 0 \\ -2 & 9-\lambda & 0 \\ 5 & 9 & 3- \lambda \end{array}\right|= 0$. I notice that the last column has '0's every where except at the bottom so we can "expand" on the final column to get
$\displaystyle (3- \lambda)\left|\begin{array}{cc} 6-\lambda & -2 \\ -2 & 9- \lambda\end{array}\right|= (3- \lambda)((6- \lambda)(9- \lambda)- 4)= (3- \lambda)(54- 15\lambda+ \lambda^2- 4)= 150- 95\lambda+ 18\lambda^2- \lambda^3= 0$.
Not for finding the characteristic equation. The characteristic equation for your reduced matrix is $\displaystyle \left|\begin{array}{ccc}1- \lambda & \frac{1}{3} & 0 \\ 0 & 1- \lambda & 0 \\ 0 & 0 & 1- \lambda \end{array}\right|= (1- \lambda)^3$ but that is NOT the characteristic equation of the original matrix.