Skip to main contentSkip to solution

If AT=[abcd]A^T = \begin{bmatrix} a & b \\ c & d \end{bmatrix} and A−1=[pqrs]A^{-1} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}, then

Solution

✅ Correct Option: 4

Inverse and transpose commute: (AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T. Since AT=[abcd]A^T = \begin{bmatrix} a & b \\ c & d \end{bmatrix} and A−1=[pqrs]A^{-1} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}, we get [abcd]−1=[pqrs]T\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}^T.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question