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If matrix A=[p34p]A = \begin{bmatrix} p & -3 \\ -4 & p \end{bmatrix} and A3=64|A^3| = 64, then the value of p is:

Solution

Correct Option: 4

When a matrix is raised to a power, the determinant is also raised to that same power.

A3=A3|A^3| = |A|^3


Given that A3=64|A^3| = 64:

A3=64|A|^3 = 64

A=643|A| = \sqrt[3]{64}

A=4|A| = 4


For a 2×2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is adbcad - bc.

For matrix A=[p34p]A = \begin{bmatrix} p & -3 \\ -4 & p \end{bmatrix}:

A=(p)(p)(3)(4)|A| = (p)(p) - (-3)(-4)

A=p212|A| = p^2 - 12


Since A=4|A| = 4:

p212=4p^2 - 12 = 4

p2=16p^2 = 16

p=±4p = \pm 4

Therefore, the value of pp is ±4\pm 4.

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