Question

Let \( A \) be a \( 3 \times 3 \) real matrix such that**
\(A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}.\)
Then, the system \( (A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) has

• A. unique solution
• B. exactly two solutions
• C. no solution
• D. infinitely many solutions

Answer 1

The Correct Option is A

To solve the problem, let's first analyze the given information about the matrix \( A \) and its actions on specific vectors:

1. \( A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \), which implies that \(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\) is an eigenvector of \( A \) with eigenvalue 2.
2. \( A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \).
3. \( A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \), which implies that \(\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\) is also an eigenvector of \( A \) with eigenvalue 2.

Now, let's form the matrix equation for vector multiplication:

Using the information from above, the general form of \( A \) can be constructed since the eigenvectors associated with eigenvalues will contribute to columns:

Let \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \). Substituting the given vectors, we have:

From \( A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \):

• \( a + c = 2 \),
• \( d + f = 0 \),
• \( g + i = 2 \).

From \( A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \):

• \( b + c = -4 \),
• \( e + f = 0 \),
• \( h + i = 4 \).\li>

From \( A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \):

• \( a + b = 2 \),
• \( d + e = 2 \),
• \( g + h = 0 \).

These equations ensure the structure of matrix \( A \) with specific row operations and eigenvectors confirmed.

The system provided is \((A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \).

Since we know the eigenvalues are 2, 4, and from further analysis:

• If \( \lambda = 2 \), then suggestively, the reduced matrix has a determinant or rank leading to unique solutions based on rank-nullity concept.
• The eigenvalue 3 isn't one due to matrix implications that implies invertibility being possible with uniquely solvable system via Cramer’s Rule having an existent determinant of nonzero value.

As for the provided options:

1. Unique Solution: Justified as with slight analysis of vector eigen-transformation which disrupts only due to matrix order.
2. Exactly two solutions, No solution, Infinitely many solutions: Each is incorrect with only unique remuneration due to above logic of matrix properties.

Thus, the system has a unique solution.

Answer 2

Define the matrix \( A \) with elements. Let

\[ A = \begin{pmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{pmatrix}. \]

Use the given conditions to form equations. Using the dot product notation for matrix multiplication, we have the following conditions:

From

\[ A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} : \]

\[ (x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 0, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 1. \]

Expanding each dot product:

\[ x_1 + z_1 = 2, \quad x_2 + z_2 = 0, \quad x_3 + z_3 = 1. \quad (1) \]

From

\[ A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} : \]

\[ (x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 4, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 1. \]

Expanding each dot product:

\[ x_1 + y_1 + z_1 = 4, \quad x_2 + y_2 + z_2 = 0, \quad x_3 + y_3 + z_3 = 1. \quad (2) \]

From

\[ A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} : \]

\[ (x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 2, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 1, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 0. \]

Expanding each dot product:

\[ y_1 + z_1 = 2, \quad y_2 + z_2 = 1, \quad y_3 + z_3 = 0. \quad (3) \]

Solve for elements of \( A \). Using equations (1), (2), and (3), we can solve for the individual elements of \( A \):

From equation (2): \( x_1 + y_1 + z_1 = 4 \) and \( y_1 + z_1 = 2 \) from equation (3). Substitute \( z_1 = 2 - y_1 \) into equation (1) to find \( x_1, y_1, z_1 \).

Similarly, solve for \( x_2, y_2, z_2 \) and \( x_3, y_3, z_3 \) to complete the matrix \( A \).

Set up the system:

\[ (A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}. \]

Now, calculate \( A - 3I \) and substitute to find the unique solution for the system.

Therefore, the answer is: unique solution.