(C) Given,$A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = \begin{vmatrix} \alpha & 2 \\

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(C) Given,$A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = \begin{vmatrix} \alpha & 2 \\ 2 & \alpha \end{vmatrix} = \alpha^{2} - 4 \quad (i)$.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:$|A| = \sqrt[3]{125} = 5$.Now,substitute $|A| = 5$ into equation $(i)$:$5 = \alpha^{2} - 4$.$\alpha^{2} = 5 + 4 = 9$.Taking the square root on both sides:$\alpha = \pm 3$.

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 $\alpha$ is equal to

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$\pm 1$
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$\pm 2$
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$\pm 3$
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$\pm 5$

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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 $\alpha$ is equal to

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The value of $a$ for which the system of equations has a non-zero solution is
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$