I need to find the eigenvalue of the following matrix (1):
$$A = \begin{bmatrix}
2 & -1 & 0 \\
-1 & 3 & 0 \\
0 & 0 & 7
\end{bmatrix}$$
for this I need to compute (2)
$$\det{A – \lambda I} = \det\Big(\begin{bmatrix}
2-\lambda & -1 & 0 \\
-1 & 3-\lambda & 0 \\
0 & 0 & 7-\lambda
\end{bmatrix}\Big)$$
which can be developped in (3) which is the correct answer given
$$(\lambda^{2} -4\lambda + 3)(7-\lambda)$$
However if I follow the algorithm to determine the determinant of a 3×3 matrix (4)
$$\text{if} \quad A = \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}, \quad \text{then} \quad \det(A) = a\begin{bmatrix}
e & f \\
h & i
\end{bmatrix} – b\begin{bmatrix}
d & f \\
g & i
\end{bmatrix} +c\begin{bmatrix}
d & e \\
g & h
\end{bmatrix}$$
What I would get is (5)
$$(2-\lambda)(2-\lambda)(7-\lambda)-(-1)(-1)(7-\lambda) = (4-4\lambda + \lambda^{2})(7-\lambda) -(7-\lambda)$$
What I don't understand is how to get to the equation (3)?
Best Answer
Use the fact that$$(4-4\lambda+\lambda^2)(7-\lambda)-(7-\lambda)=\bigl((4-4\lambda+\lambda^2)-1)(7-\lambda)=(3-4\lambda+\lambda^2)(7-\lambda).$$