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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].
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].
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