Question:
If
\[
A=\begin{bmatrix}
x & 2 & 1\\
2 & x & 1\\
2 & 1 & 0
\end{bmatrix}
\]
and \(\det(A^3)=125\), then \(x=\)
If
\[
A=\begin{bmatrix}
x & 2 & 1\\
2 & x & 1\\
2 & 1 & 0
\end{bmatrix}
\]
and \(\det(A^3)=125\), then \(x=\)
Show Hint
Use the determinant property \(\det(A^n)=(\det A)^n\) before expanding the determinant.
Updated On: Jun 26, 2026
- \(\frac{1}{3}\)
- \(3\)
- \(-\frac{1}{3}\)
- \(-3\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use determinant property.
We know that \[ \det(A^3)=(\det A)^3 \] Given, \[ \det(A^3)=125 \] So, \[ (\det A)^3=125 \] Therefore, \[ \det A=5 \]
Step 2: Find \(\det A\).
\[ A=\begin{bmatrix} x & 2 & 1\\ 2 & x & 1\\ 2 & 1 & 0 \end{bmatrix} \] Expanding along the first row, \[ \det A=x \begin{vmatrix} x & 1\\ 1 & 0 \end{vmatrix} -2 \begin{vmatrix} 2 & 1\\ 2 & 0 \end{vmatrix} +1 \begin{vmatrix} 2 & x\\ 2 & 1 \end{vmatrix} \] \[ =x(0-1)-2(0-2)+(2-2x) \] \[ =-x+4+2-2x \] \[ =6-3x \]
Step 3: Equate determinant with 5.
Since \[ \det A=5 \] we get \[ 6-3x=5 \] \[ -3x=-1 \] \[ x=\frac{1}{3} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{3}} \]
We know that \[ \det(A^3)=(\det A)^3 \] Given, \[ \det(A^3)=125 \] So, \[ (\det A)^3=125 \] Therefore, \[ \det A=5 \]
Step 2: Find \(\det A\).
\[ A=\begin{bmatrix} x & 2 & 1\\ 2 & x & 1\\ 2 & 1 & 0 \end{bmatrix} \] Expanding along the first row, \[ \det A=x \begin{vmatrix} x & 1\\ 1 & 0 \end{vmatrix} -2 \begin{vmatrix} 2 & 1\\ 2 & 0 \end{vmatrix} +1 \begin{vmatrix} 2 & x\\ 2 & 1 \end{vmatrix} \] \[ =x(0-1)-2(0-2)+(2-2x) \] \[ =-x+4+2-2x \] \[ =6-3x \]
Step 3: Equate determinant with 5.
Since \[ \det A=5 \] we get \[ 6-3x=5 \] \[ -3x=-1 \] \[ x=\frac{1}{3} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{3}} \]
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