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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    Let A and B be two invertible matrices of order \[3\times 3.\]If \[\det (AB{{A}^{T}})=8\]and \[\det (A{{B}^{-1}})=8,\]then \[\det (B{{A}^{-1}}{{B}^{T}})\]is equal to [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 1                                 

    B) 16    

    C) 1/16 

    D)   ¼

    Correct Answer: C

    Solution :

    Here, \[|A{{|}^{2}}.|B|=8\]               ...(i) and \[\frac{|A|}{|B|}=8\]                                ...(ii) Solving (i) and (ii), we get\[|A|=4\,\text{and}|B|\,=\,\frac{1}{2}\] Now, det\[(B{{A}^{-1}}{{B}^{T}})\,=\,\frac{|B{{|}^{2}}}{|A|}=\frac{1}{4}.\frac{1}{4}=\frac{1}{16}\]                


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