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MHT CET20255 May 2025Evening ShiftMathematicsMatricesActual

If A= [ array rrr 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 array ] then A(I+ adj A)=

Options

  1. A[ array rrr 9 & -2 & 2 0 & 10 & -3 3 & -2 & 11 array ]
  2. B[ array ccr 8 & -2 & 2 0 & 9 & -3 3 & -2 & 10 array ]
  3. C[ array rrr 9 & -2 & 2 0 & 10 & -3 3 & -2 & 12 array ]
  4. D[ array crr 3 & 2 & -2 0 & 10 & 3 -3 & 2 & 12 array ]

Correct answer

C. [ array rrr 9 & -2 & 2 0 & 10 & -3 3 & -2 & 12 array ]

Step-by-step solution

Given A(I + adj A) , expand to AI + A( adj A) . Since AI = A and A( adj A) = |A|I , the expression simplifies to A + |A|I . Compute the determinant of A = bmatrix 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 bmatrix : |A| = 1 vmatrix 2 & -3 -2 & 4 vmatrix - (-2) vmatrix 0 & -3 3 & 4 vmatrix + 2 vmatrix 0 & 2 3 & -2 vmatrix |A| = 1(8 - 6) + 2(0 - (-9)) + 2(0 - 6) = 2 + 18 - 12 = 8 Substitute to get A + 8I = bmatrix 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 bmatrix + bmatrix 8 & 0 & 0 0 & 8 & 0 0 & 0 & 8 bmatrix Add the matrices element-

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