Question

Let A be a \(3×3\) real matrix such that \(A\begin{pmatrix}   1 \\1  \\ 0  \end{pmatrix}\)=\(\begin{pmatrix}   1 \\1  \\ 0  \end{pmatrix}\); \(A\begin{pmatrix}   1 \\0  \\ 1  \end{pmatrix}\)=\(A\begin{pmatrix}   -1 \\0  \\ 1  \end{pmatrix}\) and \(A\begin{pmatrix}   0 \\0  \\ 1  \end{pmatrix}\)=\(\begin{pmatrix}   1 \\1  \\ 2  \end{pmatrix}\)
If \(X = [x_1, x_2, x_3]^T \)and \(I\) is an identity matrix of order \(3\), then the system \([A−2I]X \)= \(\begin{pmatrix}   4 \\1  \\ 1 \end{pmatrix}\) has:

• A. No solution
• B. Infinitely many solutions
• C. Unique solution
• D. Exactly two solutions

Answer 1

The Correct Option is B

To solve the given problem, we need to analyze the properties of the given matrix \( A \) based on the system of equations provided. We are required to determine the number of solutions for the system \([A - 2I]X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}\).

Given transformation equations involving matrix \( A \):

• \( A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \)
• \( A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = A \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \)
• \( A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \)

Let's decipher what these equations signify:

The first equation implies that \(\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) is an eigenvector of \( A \) associated with eigenvalue 1.

The second equation implies that the images of \(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\) under \( A \) are identical. Therefore, their difference should map to the zero vector under \( A \), suggesting that \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) lies in the null space of \( A \). This means:

• \( A \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)

The third equation gives another transformation under \( A \).

Now, let's construct a potential matrix \( A \):

Taking these into account, one possible form of \( A \) considering transformations is:

• \( A = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 0 \end{pmatrix} \)

Next, we solve for \( [A - 2I] \):

\([A - 2I] = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & 0 \end{pmatrix} - 2\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 2 & -2 \end{pmatrix}\)

Finding the rank of \([A - 2I]\):

The rows of \([A - 2I]\) do not span all three dimensions, implying that the matrix is singular. Thus, \([A - 2I]\) has a rank of less than 3, and its determinant is zero.

Now, examine the system for solutions:

The system \([A - 2I]X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}\) implies a linear dependence in the system. Given the rank deficiency, this matrix suggests a homogeneous component, equivalent to having infinitely many solutions.

Conclusion: Since the rank of \([A - 2I]\) is less than 3, the rank does not equal the augmented matrix which indicates that the system \([A - 2I]X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}\) has infinitely many solutions.