(A) Given $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = x(0 - 1) - 2(0 - 2) +

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(A) Given $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = x(0 - 1) - 2(0 - 2) + 1(2 - 2x)$$|A| = -x + 4 + 2 - 2x = 6 - 3x$.We are given $\det(A^3) = 125$.Using the property $|A^n| = |A|^n$,we have $|A|^3 = 125$.Taking the cube root on both sides,$|A| = 5$.Substitute the value of $|A|$: $6 - 3x = 5$.$3x = 6 - 5 = 1$.$x = 1/3$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$ and $\det(A^3) = 125$,then $x =$

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$1/3$
B
$3$
C
$-1/3$
D
$-3$

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$ and $\det(A^3) = 125$,then $x =$

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