Question

If \( A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 8 & 0 \\7 & 1 \end{bmatrix} \), and \( A^3 = B \), then \( x = \) ?

• A. \(-2\) or \(3\)
• B. \(-2\)
• C. \(2\) or \(-3\)
• D. \(2\)

Hint:

To verify matrix power relationships, compute \( A^2 \) and \( A^3 \) step-by-step. Equate corresponding elements with the target matrix to solve for unknowns.

Answer 1

The Correct Option is D

We are given: \[ A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 8 & 0\\ 7 & 1 \end{bmatrix} \] and \[ A^3 = B \] First, compute \( A^2 \): \[ A^2 = A \cdot A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \\ x + 1 & 1 \end{bmatrix} \] Next, compute \( A^3 = A . A^2 \): \[ A^3 = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x^2 & 0 \\ x+1 & 1 \end{bmatrix} = \begin{bmatrix} x \cdot x^2 + 0 \cdot (x+1) & x \cdot 0 + 0 \cdot 1 \\ 1 \cdot x^2 + 1 \cdot (x+1) & 1 \cdot 0 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} x^3 & 0\\ x^2 + x + 1 & 1 \end{bmatrix} \] Now, equate \( A^3 \) with \( B \): \[ \begin{bmatrix} x^3 & 0 \\ x^2 + x + 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix} \] From the first row, first column: \[ x^3 = 8 \Rightarrow x = 2 \] Also verify the second row, first column: \[ x^2 + x + 1 = 4 + 2 + 1 = 7 \] So, \( x = 2 \) satisfies both.