For i=1,2,3 and j=1,2,3. If ai2+bi2+ci2=1, a_ia_j+b_ib_j+c_ic_j=0,∀i≠j and A= a_1 a_2 a_3 b_1 b_2 b_3 c_1 c_2 c_3 then detAA^T=

Correct answer: (B) 1. We know AA^T=AA^T=A^2 i.e. a1a2a3b1b2b3c1c2c3^2= a_1 b_1 c_1 a_2 b_2 c_2 a_3 b_3 c_3 a_1 b_1 c_1 a_2 b_2 c_2 a_3 b_3 c_3 = Σa12 Σa_1a_2 Σa_1a_3 Σa_2a_1 Σa22 Σa_2a_3 Σa_3a_1 Σa_3a_2 Σa32 = 1 0 0 0 1 0 0 0 1 =1

Q43 / 160 · AP EAMCET 2022 (04 Jul Shift 1) · Mathematics

For $i = 1, 2, 3$ and $j = 1, 2, 3$ . If $a_{i}^{2} + b_{i}^{2} + c_{i}^{2} = 1, a_{i} a_{j} + b_{i} b_{j} + c_{i} c_{j} = 0, \forall i \neq j$ and $A = [\begin{array}{l} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}]$ then $det (A A^{T}) =$

A
$0$
Correct answer
B
$1$
C
$- 1$
D
$3$

Solution

We know $|A A^{T}| = |A| |A^{T}| = {|A|}^{2}$
i.e. ${|\begin{array}{l} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}|}^{2} = |\begin{array}{l} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}| |\begin{array}{l} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}|$
$= |\begin{array}{l} Σ a_{1}^{2} & Σ a_{1} a_{2} & Σ a_{1} a_{3} \\ Σ a_{2} a_{1} & Σ a_{2}^{2} & Σ a_{2} a_{3} \\ Σ a_{3} a_{1} & Σ a_{3} a_{2} & Σ a_{3}^{2} \end{array}| = |\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}|$
$= 1$

Hence, the correct answer is option: 'B'.

For i=1,2,3 and j=1,2,3. If ai2+bi2+ci2=1, aiaj+bibj+cicj=0,∀i≠j and A=a1a2a3b1b2b3c1c2c3 then detAAT=

Solution

Relevant Questions

For $i = 1, 2, 3$ and $j = 1, 2, 3$ . If $a_{i}^{2} + b_{i}^{2} + c_{i}^{2} = 1, a_{i} a_{j} + b_{i} b_{j} + c_{i} c_{j} = 0, \forall i \neq j$ and $A = [\begin{array}{l} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}]$ then $det (A A^{T}) =$