Answer :

B =

We need to find B^{-1}

To find the inverse of a matrix we need to find the Adjoint of that matrix

For finding the adjoint of the matrix we need to find its cofactors

Let B_{ij} denote the cofactors of Matrix B

Minor of an element b_{ij} = M_{ij �}

b_{11} = 1, Minor of element b_{11} = M_{11} = = (3 × 1) – ((-2) × 0) = 3

b_{12} = 2, Minor of element b_{12} = M_{12} = = ((-1) × 1) – (0 × 0) = -1

b_{13} = -2, Minor of element b_{13} = M_{13} = = ((-1) × (-2)) – (3 × 0) = 2

b_{21} = -1, Minor of element b_{21} = M_{21} = = (2 × 1) – ((-2) × (-2)) = -2

b_{22} = 3, Minor of element b_{22} = M_{22} = = (1 × 1) – ((-2) × 0) = 1

b_{23} = 0, Minor of element b_{23} = M_{23} = = (1 × (-2)) – (2 × 0) = -2

b_{31} = 0, Minor of element b_{31} = M_{31} = = (2 × 0) – ((-2) × 3) = 6

b_{32} = -2, Minor of element b_{32} = M_{32} = = (1 × 0) – ((-2) × (-1)) = -2

b_{33} = 1, Minor of element b_{33} = M_{33} = = (1 × 3) – (2 × (-1)) = 5

Cofactor of an element b_{ij}, B_{ij} = (-1)^{i+j} × M_{ij}

B_{11} = (-1)^{1+1}× M_{11} = 1 × 3 = 3

B_{12} = (-1)^{1+2}× M_{12} = (-1) × (-1) = 1

B_{13} = (-1)^{1+3}× M_{13} = 1 × 2 = 2

B_{21} = (-1)^{2+1}× M_{21} = (-1) × (-2) = 2

B_{22} = (-1)^{2+2} × M_{22} = 1 × 1 = 1

B_{23} = (-1)^{2+3} × M_{23} = (-1) × (-2) = 2

B_{31} = (-1)^{3+1} × M_{31} = 1 × 6 = 6

B_{32} = (-1)^{3+2} × M_{32} = (-1) × (-2) = 2

B_{33} = (-1)^{3+3} × M_{33} = 1 × 5 = 5

Adj B = =

|B| = 1 (3 × 1 – (-2) × 0) – 2 ((-1) × 1 – 0 × 0) + (-2) ((-1) × (-2) – 3 × 0)

|B| = 3 – 2 ( -1 – 0) – 2 (2 – 0)

|B| = 3 + 2 – 4 = 1

∴ B^{-1} = (Adj B)/|B| = /1 =

We know (AB)^{-1} = B^{-1}A^{-1}

(AB)^{-1} = ×

Solving the above matrix we get

(AB)^{-1} =

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