To find the values of [tex]\(\lambda\)[/tex] for which the matrix [tex]\(\lambda I - A\)[/tex] is singular, we need to follow these steps:
1. Given Matrix [tex]\(A\)[/tex] and Identity Matrix [tex]\(I\)[/tex]:
We start with the matrix [tex]\(A\)[/tex]:
[tex]\[
A = \begin{pmatrix}
1 & 0 & 2 \\
0 & -1 & -2 \\
2 & -2 & 0 \\
\end{pmatrix}
\][/tex]
The identity matrix [tex]\(I\)[/tex] of the same size (3x3) is:
[tex]\[
I = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
\][/tex]
2. Form the Matrix [tex]\(\lambda I - A\)[/tex]:
The matrix [tex]\(\lambda I - A\)[/tex] is calculated as follows:
[tex]\[
\lambda I - A = \lambda \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
-
\begin{pmatrix}
1 & 0 & 2 \\
0 & -1 & -2 \\
2 & -2 & 0 \\
\end{pmatrix}
\][/tex]
Simplifying this, we get:
[tex]\[
\lambda I - A = \begin{pmatrix}
\lambda - 1 & 0 & -2 \\
0 & \lambda + 1 & 2 \\
-2 & 2 & \lambda \\
\end{pmatrix}
\][/tex]
3. Determinant of [tex]\(\lambda I - A\)[/tex]:
To find the values of [tex]\(\lambda\)[/tex] that make this matrix singular, we need to set the determinant of [tex]\(\lambda I - A\)[/tex] to zero and solve for [tex]\(\lambda\)[/tex].
So, we compute the determinant:
[tex]\[
\det(\lambda I - A) = \begin{vmatrix}
\lambda - 1 & 0 & -2 \\
0 & \lambda + 1 & 2 \\
-2 & 2 & \lambda \\
\end{vmatrix}
\][/tex]
4. Calculate the Determinant:
We expand the determinant along the first row:
[tex]\[
\det(\lambda I - A) = (\lambda - 1) \begin{vmatrix}
\lambda + 1 & 2 \\
2 & \lambda \\
\end{vmatrix}
- 0 \cdot \begin{vmatrix}
0 & -2 \\
-2 & \lambda \\
\end{vmatrix}
- 2 \begin{vmatrix}
0 & \lambda + 1 \\
-2 & 2 \\
\end{vmatrix}
\][/tex]
Further simplifying each term:
[tex]\[
(\lambda - 1) \left( (\lambda + 1)\lambda - (2 \cdot 2) \right)
- 2 \left( 0 \cdot 2 - (-2) \cdot (\lambda + 1) \right)
\][/tex]
[tex]\[
= (\lambda - 1)(\lambda^2 + \lambda - 4) + 4(\lambda + 1)
\][/tex]
Expanding and combining like terms:
[tex]\[
(\lambda - 1)\lambda^2 + (\lambda - 1)\lambda - 4(\lambda - 1) + 4\lambda + 4
\][/tex]
[tex]\[
= \lambda^3 - \lambda^2 + \lambda^2 - \lambda - 4\lambda + 4 + 4\lambda + 4
\][/tex]
Simplifying, we have:
[tex]\[
\lambda^3 - 9\lambda
\][/tex]
Hence, the determinant is:
[tex]\[
\det(\lambda I - A) = \lambda^3 - 9\lambda
\][/tex]
5. Solve for [tex]\(\lambda\)[/tex]:
Set the determinant to zero:
[tex]\[
\lambda^3 - 9\lambda = 0
\][/tex]
Factorize the expression:
[tex]\[
\lambda (\lambda^2 - 9) = 0
\][/tex]
[tex]\[
\lambda (\lambda - 3)(\lambda + 3) = 0
\][/tex]
Solving for [tex]\(\lambda\)[/tex], we get:
[tex]\[
\lambda = 0, \, \lambda = 3, \, \lambda = -3
\][/tex]
Therefore, the values of [tex]\(\lambda\)[/tex] for which the matrix [tex]\(\lambda I - A\)[/tex] is singular are [tex]\(\boxed{-3, 0, \text{ and } 3}\)[/tex].
