First slide

Transpose of matrix

Question

Let $A = (\begin{matrix} 0 & 2 b & c \\ a & b & - c \\ a & - b & c \end{matrix})$ be an orthogonal matrix then the values of a, b, c are

Moderate

A

$b = \pm \frac{1}{\sqrt{6}}, c = \pm \frac{1}{\sqrt{3}}$
B

$a = \pm \frac{1}{\sqrt{2}}, c = \pm \frac{1}{\sqrt{6}}$
C

$a = \pm \frac{1}{\sqrt{2}}, b = \pm \frac{1}{\sqrt{6}}$
D

All of these

Solution

$A = (\begin{matrix} 0 & 2 b & c \\ a & b & - c \\ a & - b & c \end{matrix}) a n d A' = (\begin{matrix} 0 & a & a \\ 2 b & b & - b \\ c & - c & c \end{matrix})$
As A is orthogonal
$∴$ AA′ = I
$\Rightarrow (\begin{matrix} 0 & 2 b & c \\ a & b & - c \\ a & - b & c \end{matrix}) (\begin{matrix} 0 & a & a \\ 2 b & b & - b \\ c & - c & c \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})$
$\Rightarrow (\begin{matrix} 4 b^{2} + c^{2} & 2 b^{2} - c^{2} & - 2 b^{2} + c^{2} \\ 2 b^{2} - c^{2} & a^{2} + b^{2} + c^{2} & a^{2} - b^{2} - c^{2} \\ - 2 b^{2} + c^{2} & a^{2} - b^{2} - c^{2} & a^{2} + b^{2} + c^{2} \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})$
$∴$ Using definition of equality of two matrices
$4 b^{2} + c^{2} = 1, 2 b^{2} - c^{2} = 0, a^{2} + b^{2} + c^{2} = 1$
On solving them,
$a = \pm \frac{1}{\sqrt{2}}, b = \pm \frac{1}{\sqrt{6}}, c = + \frac{1}{\sqrt{3}}$

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