Let A- 0 2q r p q -r p -q r - If AAT-I3- then -P- is-

The correct option is B 1√(2) A is orthogonal matrix, since #160; #160; AA^T=I_3 ⇒ 0^2+p^2+p^2=1⇒|p|=1√(2) #160; ...

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Continuity in an Interval

Let A= 0 ...

Question

Let $A = ⎛ ⎜ ⎝ \begin{matrix} 0 & 2 q & r p & q & - r p & - q & r \end{matrix} ⎞ ⎟ ⎠$ . If $A A^{T} = I_{3},$ then $| P |$ is?

\frac{1}{\sqrt{2}}

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\frac{1}{\sqrt{5}}

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\frac{1}{\sqrt{6}}

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\frac{1}{\sqrt{3}}

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Solution

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A = ⎛ ⎜ ⎝ \begin{matrix} 0 & 2 q & r p & q & - r p & - q & r \end{matrix} ⎞ ⎟ ⎠

A A^{T} = I_{3}

| p |

A = ⎛ ⎜ ⎝ \begin{matrix} 0 & 2 q & r p & q & - r p & - q & r \end{matrix} ⎞ ⎟ ⎠

A A^{T} = I_{3}

| p |

p, q

r

y^{3} - 3 y^{2} + 6 y + 1 = 0

(p q, \frac{1}{p q}),

(q r, \frac{1}{q r})

(r p, \frac{1}{r p}),

If $p > q > 0 a n d p r < - 1 < q r$ , then find the value of

{tan}^{- 1} \frac{p - q}{1 + p q} + {tan}^{- 1} \frac{q - r}{1 + q r} + {tan}^{- 1} \frac{r - p}{1 + r p} .

$I f P (Q - r) x^{2} + Q (r - P) x + r (P - Q) = 0 h a s e q u a l r o o t s t h e n \frac{2}{Q} = (w h e r e P, Q, r ϵ R)$

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