Q46 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

Consider the simultaneous linear equations $\mathrm{AX}=\mathrm{B}$ and $\mathrm{AY}=\mathrm{Q}$. If $\mathrm{A}$ is an invertible matrix and $\mathrm{B}$ is the unique solution of $\mathrm{AY}=\mathrm{Q}$, then the solution of $\mathrm{AX}=\mathrm{B}$ is

  • A
    $A^{-1}(B+Q)$
  • B
    $\left(A^{-1}\right)^2 B$
  • C
    $\mathrm{A}^{-1} \mathrm{BQ}$
  • Correct answer
    D
    $\left(A^{-1}\right)^2 Q$

Solution

Given linear equations $\mathrm{AX}=\mathrm{B}$ and $\mathrm{AY}=\mathrm{Q}$. Here, $\mathrm{AY}=\mathrm{Q}$ has unique solution $\mathrm{B}$. Then, $\mathrm{AB}=\mathrm{Q}$ Take, $\mathrm{AY}=\mathrm{B}$ Multiply both side by A. $\begin{aligned} & \mathrm{A}^2 \mathrm{X}=\mathrm{AB} \\ & \mathrm{A}^2 \mathrm{X}=\mathrm{Q} \end{aligned}$ \{Given\} Now, $\mathrm{A}$ is invertible then multiply both sides by $\mathrm{A}^{-2}$. $\left(A^2 \cdot A^{-2}\right) . X=\left(A^{-1}\right)^2 Q$ So, $X=\left(A^{-1}\right)^2 Q$ Option (d) is correct.

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

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