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 \(x=\)
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 \(x=\)
Show 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.
Updated On: Jun 24, 2026
- \(-2\) or \(3\)
- \(-2\)
- \(2\) or \(-3\)
- \(2\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Compute \(A^2\).
Given, \[ A= \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \] Now, \[ A^2=A\cdot A \] \[ A^2= \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \] Multiplying, \[ A^2= \begin{bmatrix} x^2 & 0\\ x+1 & 1 \end{bmatrix} \]
Step 2: Compute \(A^3\).
Now, \[ A^3=A^2\cdot A \] \[ A^3= \begin{bmatrix} x^2 & 0\\ x+1 & 1 \end{bmatrix} \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \] Multiplying, \[ A^3= \begin{bmatrix} x^3 & 0\\ x^2+x+1 & 1 \end{bmatrix} \]
Step 3: Compare \(A^3\) with \(B\).
Given, \[ A^3=B= \begin{bmatrix} 8 & 0\\ 7 & 1 \end{bmatrix} \] Comparing corresponding entries, \[ x^3=8 \] Thus, \[ x=2 \] Now check the second entry: \[ x^2+x+1=7 \] Substituting \(x=2\), \[ 2^2+2+1=4+2+1 \] \[ =7 \] Hence, the value satisfies both conditions.
Step 4: Final conclusion.
Therefore, \[ \boxed{2} \]
Given, \[ A= \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \] Now, \[ A^2=A\cdot A \] \[ A^2= \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \] Multiplying, \[ A^2= \begin{bmatrix} x^2 & 0\\ x+1 & 1 \end{bmatrix} \]
Step 2: Compute \(A^3\).
Now, \[ A^3=A^2\cdot A \] \[ A^3= \begin{bmatrix} x^2 & 0\\ x+1 & 1 \end{bmatrix} \begin{bmatrix} x & 0\\ 1 & 1 \end{bmatrix} \] Multiplying, \[ A^3= \begin{bmatrix} x^3 & 0\\ x^2+x+1 & 1 \end{bmatrix} \]
Step 3: Compare \(A^3\) with \(B\).
Given, \[ A^3=B= \begin{bmatrix} 8 & 0\\ 7 & 1 \end{bmatrix} \] Comparing corresponding entries, \[ x^3=8 \] Thus, \[ x=2 \] Now check the second entry: \[ x^2+x+1=7 \] Substituting \(x=2\), \[ 2^2+2+1=4+2+1 \] \[ =7 \] Hence, the value satisfies both conditions.
Step 4: Final conclusion.
Therefore, \[ \boxed{2} \]
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