Q44 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

Let $\mathrm{A}, \mathrm{B}$ be two $3 \times 3$ matrices and $\mathrm{C}$ be a $3 \times 3$ unit matrix such that $A B-C$ is a non-singular matrix. Let $\mathrm{D}=(\mathrm{AB}-\mathrm{C})^{-1}$. Then, consider the following statements. Statement I : $\operatorname{det}(\mathrm{BA})=\operatorname{det}(\mathrm{BA}-\mathrm{C}) \operatorname{det}(\mathrm{BDA})$ Statement II : ABD = DAB Which of the above statements is (are) true?

  • A
    Statement I is true, but Statement II is false
  • B
    Statement II is true, but Statement I is false
  • Correct answer
    C
    Both Statement I and Statement II are true
  • D
    Both Statement I and Statement II are false

Solution

Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right], B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right]$ and $C=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ (given) now $A B-C=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$ $\Rightarrow \mathrm{AB}-\mathrm{C}$ is a non-singular matrix. now $(A B-C)^{-1}=D=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$ Statement $1: \operatorname{det}(\mathrm{BA})=\operatorname{det}(\mathrm{BA}-\mathrm{C}) \operatorname{det}(\mathrm{BDA})$ $\Rightarrow 0=-1 \times 0 \Rightarrow 0=0$ hence statement- 1 is true. Statement $2: \mathrm{ABD}=\mathrm{DAB}$ $\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$ This shows statement 2 is true. Hence option (c) is True.

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

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