AP EAMCET202315 May 2023Evening ShiftMathematicsMatricesActual
Let B= ( array ccc 2 & 6 & 4 1 & 0 & 1 -1 & 1 & -1 array ) and C= ( array ccc -1 & 0 & 1 1 & 1 & 3 2 & 0 & 2 array ) If a matrix A is such that BAC = I , then A ⁻¹=
Options
- A( array ccc -3 & -5 & 5 0 & 9 & 14 2 & 2 & 6 array )
- B( array ccc -3 & -5 & 5 0 & 0 & 9 2 & 14 & 16 array )
- C( array ccc -3 & -5 & -6 0 & 9 & 2 2 & 14 & 6 array )
- D( array ccc -3 & -5 & -5 0 & 9 & 2 2 & 14 & 6 array )
Correct answer
D. ( array ccc -3 & -5 & -5 0 & 9 & 2 2 & 14 & 6 array )
Step-by-step solution
aligned & Given BAC =1 & (B A)⁻¹=C & A⁻¹ B⁻¹=C A⁻¹=C B & A⁻¹= [ array ccc -1 & 0 & 1 1 & 1 & 3 2 & 0 & 2 array ] [ array ccc 2 & 6 & 4 1 & 0 & 1 -1 & 1 & -1 array ]= [ array ccc -3 & -5 & -5 0 & 9 & 2 2 & 14 & 6 array ] aligned