(A) Given the matrix $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = \alpha^2 - (2 \times 2

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(A) Given the matrix $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = \alpha^2 - (2 \times 2) = \alpha^2 - 4$.We are given that $|A^3| = 125$.Using the property of determinants $|A^n| = |A|^n$,we have $|A|^3 = 125$.Taking the cube root on both sides,we get $|A| = \sqrt[3]{125} = 5$.Now,substitute the value of $|A|$ back into the equation:$\alpha^2 - 4 = 5$.$\alpha^2 = 9$.$\alpha = \pm 3$.Therefore,the correct option is $A$.

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

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If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 125$,then $\alpha = $ . . . . . .

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A
$\pm 3$
B
$\pm 2$
C
$\pm 1$
D
$\pm 5$

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3 and 4 .Determinants and Matrices

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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 $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 125$,then $\alpha = $ . . . . . .

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$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is: $(4 \sin \theta) x - 3y + z = 0$ $x - (6 \cos 2\theta) y + z = 0$ $3x - 12y + 4z = 0$

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$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is:
$(4 \sin \theta) x - 3y + z = 0$
$x - (6 \cos 2\theta) y + z = 0$
$3x - 12y + 4z = 0$