Question

Let \( A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \), \( B = [B_1, B_2, B_3] \), where \( B_1, B_2, B_3 \) are column matrices, and \( AB_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), \( AB_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \), \( AB_3 = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} \).
If \(\alpha = |B|\) and \(\beta\) is the sum of all the diagonal elements of B, then \(\alpha^3 + \beta^3\) is equal to _____.

Answer 1

Correct Answer: 28

Step 1. Define Matrices:  
  \( A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \), \( B = [B_1, B_2, B_3] \)

  where  
  \( B_1 = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} \), \( B_2 = \begin{bmatrix} x_2 \\ y_2 \\ z_2 \end{bmatrix} \), \( B_3 = \begin{bmatrix} x_3 \\ y_3 \\ z_3 \end{bmatrix} \).

Step 2. Equations from Matrix Multiplication:  
  - For \( AB_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), we get:  
    \(\begin{cases}        2x_1 + z_1 = 1 \\        x_1 + y_1 = 0 \\        x_1 + z_1 = 0      \end{cases}\)

  - For \( AB_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} \), we get:  
   \(\begin{cases}        2x_2 + z_2 = 2 \\        x_2 + y_2 = 3 \\        x_2 + z_2 = 0      \end{cases}\)

  - For \( AB_3 = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} \), we get:  
  \(\begin{cases}        2x_3 + z_3 = 3 \\        x_3 + y_3 = 2 \\        x_3 + z_3 = 1      \end{cases}\)

Step 3. Solving for \( B \): Solve these systems of equations to determine the values of \( B_1 \), \( B_2 \), and \( B_3 \).

Step 4. Calculate \( \alpha \) and \( \beta \):  
  - \( \alpha = |B| = 3 - \beta \) is the sum of the diagonal elements of \( B \), which is 1.

Step 5. Find \( \alpha^3 + \beta^3 \):  
  \( \alpha^3 + \beta^3 = 27 + 1 = 28 \)