Question

Which of the following is/are the eigenvector(s) for the matrix given below? \[ \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix} \]

• A. \(\begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix}\)
• B. \(\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}\)
• C. \(\begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix}\)
• D. \(\begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix}\)

Hint:

When solving for eigenvectors, remember to substitute each eigenvalue back into the equation \( (A - \lambda I)v = 0 \) and solve the system of linear equations to find the corresponding eigenvectors.

Answer 1

The Correct Option is A, C, D

Method: Direct Verification via Matrix-Vector Multiplication

Rather than solving the characteristic equation $\text{det}(A - \lambda I) = 0$, we can test each given option by computing the product $Av$ and checking if the resulting vector is parallel to $v$.

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Step 1: Verify Option (A)

Let $v_1 = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix}$. We calculate the product $Av_1$. If the resulting vector can be expressed as $\lambda \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix}$, then $v_1$ is a valid eigenvector.

The multiplication yields a vector that is indeed a scalar multiple of $v_1$, confirming Option (A) as a correct eigenvector.

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Step 2: Verify Option (C)

Let $v_2 = \begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix}$. We perform the multiplication $Av_2$.

The calculation shows that $Av_2$ is proportional to $v_2$, which satisfies the eigenvector condition. Therefore, Option (C) is a correct eigenvector.

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Step 3: Verify Option (D)

Let $v_3 = \begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix}$. We compute $Av_3$.

The output vector is a scalar multiple of $v_3$. This confirms that the vector in Option (D) is also a valid eigenvector for the matrix.

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Final Answer:

The correct eigenvectors for the matrix are:(A), (C), and (D)