Let A and B be two invertible matrices of order 3 × 3 . If det(ABA^ T )=8 and det (AB^ -1 )=8 , then det (BA^ -1 ~B^ T ) is equal to

Correct answer: C. (1)/(16) Let |A|=a,|B|=b . A^ =a|A^ -1 |=(1)/(a),|B^ |=b,|B^ -1 |=(1)/(b) A B A^ =8 |A||B| A^ -8 (1) a · b · a=8 a^ 2 b=8 A B^ -1 |=8 H||B^ -1 |=8 a, (1)/(b)=8 From (1) \& (2) a=4, b=(1)/(2) Then, |B A^ -1 B^ |=|B||A^ -1 ||B^ |=b · (1)/(a) · b= b^ 2 a =(1)/(16)

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Determinants

Let A and B be two invertible matrices of order 3 × 3 . If det(ABA^...

Let

A and B

be two invertible matrices of order

3 \times 3

. If

\operatorname{det}\left(\mathrm{ABA}^{\mathrm{T}}\right)=8 and det

\left(\mathrm{AB}^{-1}\right)=8

, then det \left(\mathrm{BA}^{-1} \mathrm{~B}^{\mathrm{T}}\right)

is equal to

Options

A $\frac{1}{4}$
B1
C $\frac{1}{16}$
D16

Correct answer

C. $\frac{1}{16}$

Step-by-step solution

Let

|A|=a,|B|=b \left.\Rightarrow \quad \mid A^{\eta}\right\rceil=a\left|A^{-1}\right|=\frac{1}{a},\left|B^{\eta}\right|=b,\left|B^{-1}\right|=\frac{1}{b} \because \quad A B A^{\eta}=8 \Rightarrow|A||B| \mid A^{\eta}-8 \ldots(1) \Rightarrow \quad a \cdot b \cdot a=8 \Rightarrow a^{2} b=8 \because \quad A B^{-1}|=8 \Rightarrow H|\left|B^{-1}\right|=8 \Rightarrow a, \frac{1}{b}=8

From (1) \& (2)

a=4, b=\frac{1}{2}

Then,

\left|B A^{-1} B^{\eta}\right|=|B|\left|A^{-1}\right|\left|B^{\eta}\right|=b \cdot \frac{1}{a} \cdot b=\frac{b^{2}}{a}=\frac{1}{16}

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