(A) Given the matrix $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$.We know that the determinant of a triangular matrix is

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(A) Given the matrix $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$.We know that the determinant of a triangular matrix is the product of its diagonal elements.Thus,$|A| = 5 \times x \times 5 = 25x$.We are given that $|A^2| = 25$.Using the property of determinants,$|A^2| = |A|^2$.Therefore,$(25x)^2 = 25$.$625x^2 = 25$.$x^2 = \frac{25}{625} = \frac{1}{25}$.Taking the square root on both sides,$|x| = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

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 $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$ and $|A^2| = 25$,then $|x|$ is equal to

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A
$\frac{1}{5}$
B
$5$
C
$25$
D
$1$

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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 $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$ and $|A^2| = 25$,then $|x|$ is equal to

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