b) Find [tex]\lambda[/tex] for which the matrix [tex]\lambda I - A[/tex] is a singular matrix, where [tex]I[/tex] is an identity matrix, given that

[tex]\[ A = \begin{pmatrix}
1 & 0 & 2 \\
0 & -1 & -2 \\
2 & -2 & 0
\end{pmatrix} \][/tex]



Answer :

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].