Chapter 5: Problem 2
Let $$ \boldsymbol{A}=\left[\begin{array}{rr} -1 & 2 \\ 4 & 1 \end{array}\right] \text { and } \quad \boldsymbol{B}=\left[\begin{array}{ll} 1 & 1 \\ \lambda & \mu \end{array}\right] $$ where \(\lambda \neq \mu\). Find all pairs of values \(\lambda, \mu\) such that \(\boldsymbol{B}^{-1} \boldsymbol{A B}\) is a diagonal matrix.
Short Answer
Step by step solution
Compute \( \boldsymbol{AB} \)
Compute \( \boldsymbol{B}^{-1} \)
Compute \( \boldsymbol{B}^{-1} \boldsymbol{A} \boldsymbol{B} \)
Equate to a Diagonal Matrix and Solve
Final Step: Conclusion of Pairs
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Multiplication
This results in a new matrix whose elements are calculated as follows:
- Take the corresponding elements from the row of the first matrix.
- Multiply these elements with the elements in the column of the second matrix.
- Sum the products to get a single number which will be an element in the resulting matrix.
Inverse Matrix
- Not every matrix has an inverse; only square matrices that are non-singular (i.e., have a non-zero determinant) do.
- The formula for finding the inverse of a 2x2 matrix \(\begin{bmatrix}x & y \ z & w \end{bmatrix}\)goes as follows: if \(det = xw - yz \eq 0\), then\[\text{Inverse} = \frac{1}{det} \begin{bmatrix} w & -y \ -z & x \end{bmatrix}\]
Diagonal Matrix
- Matrix Multiplication: Multiplying a diagonal matrix with another matrix often results in a much simpler form to compute.
- Eigenvalues and Eigenvectors: Diagonal matrices are crucial in these topics, as they directly reveal and represent the eigenvalues of a matrix.
System of Equations
- A system of linear equations can be written in the form \(AX = B\), where \(A\) is a coefficient matrix, \(X\) a column of variables, and \(B\) a column of constants.
- The main goal is to find \(X\), which requires methods like Gaussian elimination, LU decomposition, or using the inverse matrix for direct solutions when possible.
- For instance, multiply both sides of the equation by \(A^{-1}\) (if it exists) to isolate \(X\): \(X = A^{-1}B\).