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The vector e is an eigenvector of the matrix A, with corresponding eigenvalue λ, and is also an eigenvector of the matrix B, with corresponding eigenvalue μ - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1
Question 8
The vector e is an eigenvector of the matrix A, with corresponding eigenvalue λ, and is also an eigenvector of the matrix B, with corresponding eigenvalue μ. Show th... show full transcript
Worked Solution & Example Answer:The vector e is an eigenvector of the matrix A, with corresponding eigenvalue λ, and is also an eigenvector of the matrix B, with corresponding eigenvalue μ - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1
Step 1
Show that e is an eigenvector of the matrix AB with corresponding eigenvalue λμ.
Answer
To show that e is an eigenvector of the matrix AB with corresponding eigenvalue λμ, we start from the definitions of eigenvalues and eigenvectors. Given that Ae = λe and Be = μe, we can multiply both sides by B:
Thus, we have shown that e is an eigenvector of AB with eigenvalue λμ.
Step 2
State the eigenvalues of the matrix C, where C = \( \begin{pmatrix} -1 & -1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix} \) and find corresponding eigenvectors.
Answer
To find the eigenvalues of matrix C, we look for the values of λ such that det(C - λI) = 0. The eigenvalues can be calculated as:
- From the leading diagonal, we see the eigenvalues of C are: ( -1, 1, 2 ).
The corresponding eigenvectors can be determined by solving the equations:
- For λ = -1, we solve (C + I)x = 0 to find the eigenvector ( e_1 = \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} ).
- For λ = 1, we solve (C - I)x = 0 to find the eigenvector ( e_2 = \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix} ).
- For λ = 2, we solve (C - 2I)x = 0 to find the eigenvector ( e_3 = \begin{pmatrix} 1 \ -2 \ 1 \end{pmatrix} ).
Step 3
Show that \( \begin{pmatrix} 1 \\ 6 \\ 3 \end{pmatrix} \) is an eigenvector of the matrix D, where D = \( \begin{pmatrix} 1 & -1 & 1 \\ -9 & -3 & 4 \end{pmatrix} \) and state the corresponding eigenvalue.
Answer
To show that ( \begin{pmatrix} 1 \ 6 \ 3 \end{pmatrix} ) is an eigenvector of D, we compute:
which indicates that ( egin{pmatrix} 1 \ 6 \ 3 \end{pmatrix} ) is an eigenvector with an eigenvalue of -2.
Step 4
Hence state an eigenvector of the matrix CD and give the corresponding eigenvalue.
Answer
Since CD shares eigenvectors with C and D, we note that both C and D have common eigenvalues, and thus:
The eigenvector ( \begin{pmatrix} 1 \ 6 \ 3 \end{pmatrix} ) can be recognized as an eigenvector of CD. The corresponding eigenvalue can be calculated as:
Thus, the corresponding eigenvalue relating to this eigenvector is -4.