Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
2.3k views
in Mathematics by (50.1k points)
closed by

If \(\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}\) and \(\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^{2} \mathrm{~A}\), then \(\operatorname{det} \mathrm{X}\) is equal to :

(1) 243

(2) 729

(3) 27

(4) 891

1 Answer

+2 votes
by (50.3k points)
selected by
 
Best answer

Correct option is (2) 729

\(A=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right] \Rightarrow \operatorname{det}(A)=3\)

\(B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right] \Rightarrow \operatorname{det}(B)=1\)

Now 

\(\mathrm{C}=\mathrm{ABA}^{\mathrm{T}} \Rightarrow \operatorname{det}(\mathrm{C})=(\operatorname{det}(\mathrm{A}))^{2} \mathrm{x} \operatorname{det}(\mathrm{B})\)

\(|C|=9\)

Now

\(|\mathrm{X}|=\left|\mathrm{A}^{\mathrm{T}} \mathrm{C}^{2} \mathrm{~A}\right|\)

\(=\left|\mathrm{A}^{\mathrm{T}}\right||\mathrm{C}|^{2}|\mathrm{~A}|\)

\(=|\mathrm{A}|^{2}|\mathrm{C}|^{2}\)

\(=9 \times 81\)

\(=729\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...