Q44 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

If $A+2 B=\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right)$ and $2 A-B=\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right)$ then $\operatorname{Tr}[\mathrm{A}]-\operatorname{Tr}[\mathrm{B}]=$

  • A
    1
  • Correct answer
    B
    2
  • C
    3
  • D
    4

Solution

$\begin{aligned} & \text { } \mathrm{A}+2 \mathrm{~B}=\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right) ; 2 \mathrm{~A}-\mathrm{B}=\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right) \\ & \text { Now, }(\mathrm{A}+2 \mathrm{~B})+2(2 \mathrm{~A}-\mathrm{B})= \\ & \left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right)+2\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right) \\ & \Rightarrow 5 A=\left(\begin{array}{ccc}5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5\end{array}\right) \\ & \Rightarrow A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1\end{array}\right) \\ & \because A+2 B=\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right) \Rightarrow\left(\begin{array}{ccc}1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1\end{array}\right)+2 B \\ & =\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right) \\ & \Rightarrow 2 \mathrm{~B}=\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right)+\left(\begin{array}{ccc}-1 & 0 & -2 \\ -2 & +1 & -3 \\ +1 & -1 & -1\end{array}\right)=\left(\begin{array}{ccc}0 & 2 & -2 \\ 4 & -2 & 0 \\ -4 & +2 & 0\end{array}\right) \\ & \Rightarrow \mathrm{B}=\left(\begin{array}{ccc}0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0\end{array}\right) \\ & \operatorname{Tr}(A)-\operatorname{Tr}(B)=(1-1+1)-(0-1+0)=2 \\ & \end{aligned}$

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

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