First slide

Transpose of matrix

Question

If $A = [\begin{array}{ccc} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}]$ satisfies $A^{'} = A^{- 1},$ then

Moderate

A

$x = \pm 1 / \sqrt{6}, y = \pm 1 / \sqrt{6}, z = \pm 1 / \sqrt{3}$
B

$x = \pm 1 / \sqrt{2}, y = \pm 1 / \sqrt{6}, z = \pm 1 / \sqrt{3}$
C

$x = \pm 1 / \sqrt{2}, y = \pm 1 / \sqrt{6}, z = \pm 1 / \sqrt{3}$
D

$x = \pm 1 / \sqrt{2}, y = \pm 1 / 3, z = \pm 1 / \sqrt{2}$

Solution

$A^{'} = A^{- 1} \Leftrightarrow A A^{'} = I$
Now, $A A^{'} = [\begin{array}{ccc} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}] [\begin{array}{ccc} 0 & x & x \\ 2 y & y & - y \\ z & - z & z \end{array}]$
$= [\begin{array}{ccc} 4 y^{2} + z^{2} & 2 y^{2} - z^{2} & - 2 y^{2} + z^{2} \\ 2 y^{2} - z^{2} & x^{2} + y^{2} + z^{2} & x^{2} - y^{2} - z^{2} \\ - 2 y^{2} + z^{2} & x^{2} - y^{2} - z^{2} & x^{2} + y^{2} + z^{2} \end{array}]$
Thus, $A A^{'} = I$
$\begin{array}{l} \Rightarrow 4 y^{2} + z^{2} = 1, 2 y^{2} - z^{2} = 0 \\ x^{2} + y^{2} + z^{2} = 1, x^{2} - y^{2} - z^{2} = 0 \\ ∴ x = \pm 1 / \sqrt{2}, y = \pm 1 / \sqrt{6}, z = \pm 1 / \sqrt{3} \end{array}$

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