Question

Let P be a 3×3 real matrix having eigenvalues \(\lambda_1\) = 0, \(\lambda_2\) = 1 and \(\lambda_3\) = −1. Further, \(v_1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\), \(v_2=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) and \(v_3=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) are eigenvectors of the matrix P corresponding to the eigenvalues \(\lambda_1\), \(\lambda_2\) and \(\lambda_3\), respectively. Then the entry in the first row and the third column of P is

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The Correct Option is C

To solve this problem, we need to understand how eigenvectors and eigenvalues relate to the entries of a matrix. Given that \(P\) is a 3×3 matrix with known eigenvalues and eigenvectors, we can determine the elements of the matrix step by step.

The eigenvectors given are:

• \(v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) corresponding to eigenvalue \(\lambda_1 = 0\)
• \(v_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) corresponding to eigenvalue \(\lambda_2 = 1\)
• \(v_3 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) corresponding to eigenvalue \(\lambda_3 = -1\)

The relation between a matrix \(P\) and its eigenvectors \(v\) with eigenvalue \(\lambda\) is given by the equation:

\(P v = \lambda v\)

To find the matrix \(P\), we consider its columns constituted by the eigenvectors, scaled by their respective eigenvalues:

• \(P v_1 = 0 \cdot v_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
• \(P v_2 = 1 \cdot v_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\)
• \(P v_3 = -1 \cdot v_3 = \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix}\)

The matrix \(P\) can be assembled such that:

P = \(\begin{bmatrix} 0 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}\)

From this matrix, we can determine the entry in the first row and third column, which is \( -1 \).

Therefore, the correct answer is:

-1