Answer :

Minor of an element a_{ij} = M_{ij}

a_{11} = 1, Minor of element a_{11} = M_{11} = = (1 × 1) – (0 × 0) = 1

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with the determinant. Solving this we get M_{11} = 1

Similarly, finding other Minors of the determinant

a_{12} = 0, Minor of element a_{12} = M_{12} = = (0 × 1) – (0 × 0) = 0

a_{13} = 0, Minor of element a_{13} = M_{13} = = (0 × 0) - (1 × 0) = 0

a_{21} = 0, Minor of element a_{21} = M_{21} = = (0 × 1) – (0 × 0) = 0

a_{22} = 1, Minor of element a_{22} = M_{22} = = (1 × 1) – (0 × 0) = 1

a_{23} = 0, Minor of element a_{23} = M_{23} = = (1 × 0) – (0 × 0) = 0

a_{31} = 0, Minor of element a_{31} = M_{31} = = (0 × 0) – (0 × 1) = 0

a_{32} = 0, Minor of element a_{32} = M_{32} = = (1 × 0) – (0 × 0) = 0

a_{33} = 1, Minor of element a_{33} = M_{33} = = (1 × 1) – (0 × 0) = 1

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

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

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 0 = 0

A_{13} = (-1)^{1+3} × M_{13} = 1 × 0 = 0

A_{21} = (-1)^{2+1} × M_{21} = (-1) × 0 = 0

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

A_{23} = (-1)^{2+3} × M_{23} = (-1) × 0 = 0

A_{31} = (-1)^{3+1} × M_{31} = 1 × 0 = 0

A_{32} = (-1)^{3+2} × M_{32} = (-1) × 0 = 0

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

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