Question

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

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

Hint:

When two matrices are equal, their corresponding elements must be equal. After finding matrix powers, compare entries position-wise to obtain the unknown variable.

Answer 1

The Correct Option is D

Step 1: Notice the pattern for powers of $A$.
Given $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$. For a lower-triangular matrix with diagonal entries $x$ and $1$, powers are predictable. The $(1,1)$ entry of $A^n$ is always $x^n$, and the $(2,2)$ entry is always $1^n = 1$.

Step 2: Conjecture the form of $A^n$.
Let us guess $A^n = \begin{bmatrix} x^n & 0 \\ x^{n-1}+x^{n-2}+\cdots+1 & 1 \end{bmatrix}$. That off-diagonal entry sums to $\frac{x^n - 1}{x - 1}$ when $x \neq 1$.

Step 3: Verify with $n=3$.
The $(2,1)$ entry of $A^3$ would be $x^2 + x + 1$. So \[ A^3 = \begin{bmatrix} x^3 & 0 \\ x^2+x+1 & 1 \end{bmatrix} \]

Step 4: Compare with $B$.
Given $B = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$. Matching entries: $x^3 = 8 \Rightarrow x = 2$ and $x^2+x+1 = 4+2+1 = 7$. Both match.

Step 5: Check uniqueness.
$x^3 = 8$ has only one real solution: $x = 2$. No other real value works.

Step 6: State the answer.
\[ \boxed{x = 2} \]