, APM3700 Assignment 3 (COMPLETE ANSWERS)
2024 - DUE 28 August 2024 ; 100% TRUSTED
Complete, trusted solutions and explanations.
QUESTION 1 If 3 6 1 4 B , find the eigenvalues of B.
(5)
To find the eigenvalues of the matrix BBB, we need to solve the
characteristic polynomial of BBB.
Given:
3 & 6 \\ 1 & 4 \end{pmatrix} \] The characteristic polynomial \
( p(\lambda) \) is found using the formula: \[ p(\lambda) = \
det(B - \lambda I) \] where \( I \) is the identity matrix. First,
subtract \( \lambda \) times the identity matrix from \( B \): \[ B
- \lambda I = \begin{pmatrix} 3 & 6 \\ 1 & 4 \end{pmatrix} - \
begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \
begin{pmatrix} 3 - \lambda & 6 \\ 1 & 4 - \lambda \
end{pmatrix} \] Next, calculate the determinant of this matrix: \
[ \det(B - \lambda I) = \begin{vmatrix} 3 - \lambda & 6 \\ 1 & 4 -
\lambda \end{vmatrix}
The determinant of a 2×22 \times 22×2 matrix (abcd)\
begin{pmatrix} a & b \\ c & d \end{pmatrix}(acbd) is calculated
as ad−bcad - bcad−bc. Applying this to our matrix:
det(B−λI)=(3−λ)(4−λ)−(6⋅1)\det(B - \lambda I) = (3 - \lambda)(4 -
\lambda) - (6 \cdot 1)det(B−λI)=(3−λ)(4−λ)−(6⋅1)
Expand and simplify:
2024 - DUE 28 August 2024 ; 100% TRUSTED
Complete, trusted solutions and explanations.
QUESTION 1 If 3 6 1 4 B , find the eigenvalues of B.
(5)
To find the eigenvalues of the matrix BBB, we need to solve the
characteristic polynomial of BBB.
Given:
3 & 6 \\ 1 & 4 \end{pmatrix} \] The characteristic polynomial \
( p(\lambda) \) is found using the formula: \[ p(\lambda) = \
det(B - \lambda I) \] where \( I \) is the identity matrix. First,
subtract \( \lambda \) times the identity matrix from \( B \): \[ B
- \lambda I = \begin{pmatrix} 3 & 6 \\ 1 & 4 \end{pmatrix} - \
begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \
begin{pmatrix} 3 - \lambda & 6 \\ 1 & 4 - \lambda \
end{pmatrix} \] Next, calculate the determinant of this matrix: \
[ \det(B - \lambda I) = \begin{vmatrix} 3 - \lambda & 6 \\ 1 & 4 -
\lambda \end{vmatrix}
The determinant of a 2×22 \times 22×2 matrix (abcd)\
begin{pmatrix} a & b \\ c & d \end{pmatrix}(acbd) is calculated
as ad−bcad - bcad−bc. Applying this to our matrix:
det(B−λI)=(3−λ)(4−λ)−(6⋅1)\det(B - \lambda I) = (3 - \lambda)(4 -
\lambda) - (6 \cdot 1)det(B−λI)=(3−λ)(4−λ)−(6⋅1)
Expand and simplify:
