Question

if A= |x. y. z
y z x
z x y |
be a matrix such that x,y,z are positive real numbers also xyz=2 and AA^T=4I then find value of x^3+y^3+z^3​

Answer 1

SOLUTION

GIVEN

$A = \displaystyle\begin{vmatrix} x & y & z\\ y & z & x \\ z & x & y \end{vmatrix}$

be a matrix such that x,y,z are positive real numbers also xyz = 2 and

$A{A}^{T} = 4I$

TO DETERMINE

${ {x}^{3} + {y}^{3} + {z}^{3} }$

EVALUATION

Here it is given that

$A = \displaystyle\begin{vmatrix} x & y & z\\ y & z & x \\ z & x & y \end{vmatrix}$

$= x(yz - {x}^{2} ) - y( {y}^{2} - xz) + z(yx - {z}^{2} )$

$= - {x}^{3} - {y}^{3} - {z}^{3} + 3xyz$

$= - \: ( {x}^{3} + {y}^{3} + {z}^{3} - 3xyz)$

$= - ( {x}^{3} + {y}^{3} + {z}^{3} - 6) \: \: ( \because \: \: xyz = 2)$

Again

$A{A}^{T} = 4I$

Taking determinant in both sides we get

$|A| |{A}^{T}| = | 4I\: |$

$\implies |A| |A| = 64| I\: |$

$\implies {|A|}^{2} = 64$

$\implies { \bigg [ - ( {x}^{3} + {y}^{3} + {z}^{3} - 6)\bigg ]}^{2} = 64$

$\implies { \bigg [ ( {x}^{3} + {y}^{3} + {z}^{3} - 6)\bigg ]}^{2} = 64$

$\implies ( {x}^{3} + {y}^{3} + {z}^{3} - 6) = 8$

$\implies {x}^{3} + {y}^{3} + {z}^{3} = 14$

FINAL ANSWER

$\boxed{ \: \: {x}^{3} + {y}^{3} + {z}^{3} = 14 \: \: }$

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