Matrices and Determinants 1 Question 5
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5. Let $A=p \quad q \quad \rightarrow$. If $A A^{T}=I _3$, then $|p|$ is $p \quad-q$
(2019 Main, 11 Jan I)
####5. Let $A=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right]$. If $A A^T=I_3$, then $|p|$ is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed
(a) $\frac{1}{\sqrt{5}}$
(b) $\frac{1}{\sqrt{2}}$
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{1}{\sqrt{6}}$
(2019 Main, 11 Jan I)
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Answer:
Correct Answer: 5. (b)
Solution:
- $ \begin{aligned} & \text { Given, } A A^T=I \\ & \Rightarrow\left[\begin{array}{ccc} 0 & 2 q & r \\ p & q & -r \\ p & -q & r \end{array}\right]\left[\begin{array}{ccc} 0 & p & p \\ 2 q & q & -q \\ r & -r & r \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} 0+4 q^2+r^2 & 0+2 q^2-r^2 & 0-2 q^2+r^2 \\ 0+2 q^2-r^2 & p^2+q^2+r^2 & p^2-q^2-r^2 \\ 0-2 q^2+r^2 & p^2-q^2-r^2 & p^2+q^2+r^2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $
We know that, if two matrices are equal, then corresponding elements are also equal, so
$ 4 q^2+r^2=1=p^2+q^2+r^2, \quad $…(i)
$ 2 q^2-r^2=0 \Rightarrow r^2=2 q^2\quad $…(ii)
and
$ p^2-q^2-r^2=0\quad $…(iii)
Using Eqs. (ii) and (iii), we get
$ p^2=3 q^2 $
Using Eqs. (ii) and (iv) in Eq. (i), we get
$4 q^2+2 q^2 =1 $
$\Rightarrow 6 q^2 =1 $
$\Rightarrow 2 p^2 =1 \quad$ [using Eq. (iv)]
$p^2=\frac{1}{2} \Rightarrow|p| =\frac{1}{\sqrt{2}}$
