MHT CET202619 April 2026Morning ShiftMathematicsMatricesActual
The inverse of matrix bmatrix 1+pq & p & 0 q & 1+pq & p 0 & q & 1 bmatrix is ...
Options
- Abmatrix 1+pq & p & 0 q & 1+pq & p 0 & q & 1 bmatrix
- Bbmatrix 1 & p & p^2 q & 1+pq & p+p^2q q^2 & q+pq^2 & 1+pq+p^2q^2 bmatrix
- Cbmatrix 1 & -p & p^2 -q & 1+pq & -(p+p^2q) q^2 & -(q+pq^2) & 1+pq+p^2q^2 bmatrix
- Dbmatrix 1 & -p & p^2 -q & 1+pq & p+p^2q q^2 & q+pq^2 & 1+pq+p^2q^2 bmatrix
Correct answer
C. bmatrix 1 & -p & p^2 -q & 1+pq & -(p+p^2q) q^2 & -(q+pq^2) & 1+pq+p^2q^2 bmatrix
Step-by-step solution
Let the given matrix be A = bmatrix 1+pq & p & 0 q & 1+pq & p 0 & q & 1 bmatrix . The inverse of matrix A is given by A⁻¹ = 1 |A| adj (A) . First, evaluate the determinant of A : |A| = (1+pq)[(1+pq)(1) - (p)(q)] - p[q(1) - (p)(0)] + 0 |A| = (1+pq)[1 + pq - pq] - p[q] |A| = (1+pq)(1) - pq = 1 + pq - pq = 1 Since |A| = 1 , the inverse matrix is A⁻¹ = adj (A) . To find the adjoint of A , we calculate the cofactors of each element: C₁₁ = (1+pq)(1) - pq = 1 C₁₂ = -(q(1) - 0) = -q C₁₃ = q(q) - 0 = q^2 C₂₁ = -(p(1) - 0) =