Question

If a $2 \times 2$ matrix $A$ has eigenvalues 1 and 4 with the corresponding eigenvectors $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, respectively, then $A$ is _______

• A. $\begin{pmatrix} -4 & -8 \\ 5 & 9 \end{pmatrix}$
• B. $\begin{pmatrix} 9 & -8 \\ 5 & -4 \end{pmatrix}$
• C. $\begin{pmatrix} 2 & 2 \\ 1 & 3 \end{pmatrix}$
• D. $\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}$

Hint:

You can quickly verify the options using the fundamental eigenvalue properties:
1. Sum of eigenvalues = Trace of the matrix (sum of main diagonal elements). Here, $\lambda_1 + \lambda_2 = 1 + 4 = 5$.
Looking at option (D), the trace is $3 + 2 = 5$. This matches.
2. Product of eigenvalues = Determinant of the matrix. Here, $\lambda_1 \cdot \lambda_2 = 1 \times 4 = 4$.
The determinant of option (D) is $(3)(2) - (2)(1) = 6 - 2 = 4$. This also matches, confirming (D) is correct.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question asks us to reconstruct a $2 \times 2$ matrix $A$ given its eigenvalues and their corresponding eigenvectors.

Step 2: Key Formula or Approach:
We can use the matrix diagonalization theorem. Any diagonalizable matrix $A$ can be expressed as:
\[ A = S \Lambda S^{-1} \]
where $S$ is the modal matrix containing the eigenvectors as its columns, and $\Lambda$ is the diagonal matrix containing the corresponding eigenvalues along its main diagonal.

Step 3: Detailed Explanation:

• The given eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 4$.

• The corresponding eigenvectors are:
\[ v_1 = \begin{pmatrix} 1 -1 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 2 1 \end{pmatrix} \]

• Construct the modal matrix $S$ by placing the eigenvectors in its columns:
\[ S = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \]

• Construct the diagonal eigenvalue matrix $\Lambda$:
\[ \Lambda = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix} \]

• Next, compute the inverse of matrix $S$:
The determinant of $S$ is:
\[ \det(S) = (1)(1) - (2)(-1) = 1 + 2 = 3 \]
The inverse of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$:
\[ S^{-1} = \frac{1}{3} \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix} \]

• Now, calculate the matrix product $A = S \Lambda S^{-1}$. First compute the product $S \Lambda$:
\[ S \Lambda = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 1(1) + 2(0) & 1(0) + 2(4) \\ -1(1) + 1(0) & -1(0) + 1(4) \end{pmatrix} = \begin{pmatrix} 1 & 8 \\ -1 & 4 \end{pmatrix} \]

• Next, multiply $S \Lambda$ by $S^{-1}$:
\[ A = \frac{1}{3} \begin{pmatrix} 1 & 8 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix} \]
\[ A = \frac{1}{3} \begin{pmatrix} 1(1) + 8(1) & 1(-2) + 8(1) \\ -1(1) + 4(1) & -1(-2) + 4(1) \end{pmatrix} \]
\[ A = \frac{1}{3} \begin{pmatrix} 9 & 6 \\ 3 & 6 \end{pmatrix} \]

• Dividing each element of the matrix by 3:
\[ A = \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} \]

Step 4: Final Answer
The matrix $A$ is $\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}$, which corresponds to option (D).