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JEE Main & Advanced Mathematics Determinants & Matrices Question Bank Self Evaluation Test - Matrices

  • question_answer
    Let \[A+2B=\left[ \begin{matrix}    1 & 2 & 0  \\    6 & -3 & 3  \\    -5 & 3 & 1  \\ \end{matrix} \right]\] and\[2A-B=\left[ \begin{matrix}    2 & -1 & 5  \\    2 & -1 & 6  \\    0 & 1 & 2  \\ \end{matrix} \right]\], then \[\operatorname{tr}(A)  tr(B)\] is

    A) 1

    B) 3    

    C) 2

    D) 0

    Correct Answer: C

    Solution :

    [c] Here to find the value of \[tr(A)-tr(B),\] we need not to find the matrices A and B. We can find tr(A)-tr[b] using the properties of trace of matrix, i.e., \[A+2B=\left[ \begin{matrix}    1 & 2 & 0  \\    6 & -3 & 3  \\    -5 & 3 & 1  \\ \end{matrix} \right]\Rightarrow tr(A+2B)=-1(1)\] Or \[tr(A)+2tr(B)=-1\] \[\Rightarrow 2A-B=\left[ \begin{matrix}    2 & -1 & 5  \\    2 & -1 & 6  \\    0 & 1 & 2  \\ \end{matrix} \right]\] \[\Rightarrow tr(2A-B)=3\] or \[2tr(A)-tr(B)=3(2)\] Solving (1) and (2), we get tr[a] = 1 and tr[b] = -1 \[\Rightarrow tr(A)-tr(B)=2\]


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