Question

For the matrix \[ A = \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} \] which of the following statements is/are TRUE? \\

• A. $[A]{x} = {b}$ has a unique solution
• B. $[A]{x} = {b}$ does not have a unique solution
• C. $[A]$ has three linearly independent eigenvectors
• D. $[A]$ is a positive definite matrix

Hint:

A matrix with a non-zero determinant guarantees that the system of equations will have a unique solution. Also, a matrix having distinct eigenvalues implies it has linearly independent eigenvectors.

Answer 1

The Correct Option is B, C

\\ To determine whether the system has a unique solution, we need to check the determinant of the matrix $A$. If the determinant is non-zero, the system has a unique solution. - The determinant of $A$ is: \[ \text{det}(A) = \begin{vmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{vmatrix} = 1 \times \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} - (-1) \times \begin{vmatrix} -1 & -1 \\ 0 & 1 \end{vmatrix} = 1 \times (2 - 1) + 1 \times (1) = 2 \] Since the determinant is non-zero, the system $[A]{x} = {b}$ has a unique solution. \\ However, for the matrix $A$ to have linearly independent eigenvectors, the characteristic polynomial must have three distinct eigenvalues. This matrix has three linearly independent eigenvectors, thus statement (C) is also true. \\ Thus, the correct answers are (B) and (C).