Q. 8 B4.0( 17 Votes )

# Let

Answer :

A = |A| = 1 (3 × 5 – 1 × 1) – (-2) ((-2) × 5 – 1 × 1) + 1 ((-2) × 1 – 3 × 1)

|A| = (15 – 1) + 2 (-10 – 1) + (-2 – 3)

|A| = 14 – 22 – 5 = -13

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 Aij denote the cofactors of Matrix A

Minor of an element aij = Mij �

a11 = 1, Minor of element a11 = M11 = = (3 × 5) – (1 × 1) = 14

a12 = -2, Minor of element a12 = M12 = = (-2 × 5) – (1 × 1) = -11

a13 = 1, Minor of element a13 = M13 = = (-2 × 1) – (3 × 1) = -5

a21 = -2, Minor of element a21 = M21 = = ((-2) × 5) – (1 × 1) = -11

a22 = 3, Minor of element a22 = M22 = = (1 × 5) – (1 × 1) = 4

a23 = 1, Minor of element a23 = M23 = = (1 × 1) – ((-2) × 1) = 3

a31 = 1, Minor of element a31 = M31 = = (-2 × 1) – (3 × 1) = -5

a32 = 1, Minor of element a32 = M32 = = (1 × 1) – (1 × (-2)) = 3

a33 = 5, Minor of element a33 = M33 = = (1 × 3) – ((-2) × (-2)) = -1

Cofactor of an element aij = Aij

A11 = (-1)1+1× 14 = 1 × 14 = 14

A12 = (-1)1+2× (-11) = (-1) × (-11) = 11

A13 = (-1)1+3× (-5) = 1 × (-5) = -5

A21 = (-1)2+1× (-11) = (-1) × (-11) = 11

A22 = (-1)2+2 × 4 = 1 × 4 = 4

A23 = (-1)2+3 × 3 = (-1) × 3 = -3

A31 = (-1)3+1 × (-5) = 1 × (-5) = -5

A32 = (-1)3+2 × 3 = (-1) × 3 = -3

A33 = (-1)3+3 × (-1) = 1 × (-1) = -1

Adj A = = A-1 = (Adj A)/|A|

A-1 = = (ii) To find (A-1)-1 we have to find out Adj(A-1)

A-1 = |A-1| = (-1/13)3 [14 (4 × (-1) – (-3) × (-3)) – 11 (11 × (-1) – (-3) × (-5)) + (-5) (11 × (-3) – 4 × (-5))]

|A| = (-1/13)3 [14 (-4 – 9) – 11 (-11 – 15) – 5 (-33 + 20)]

|A| = (-1/13)3 [14 × (-13) – 11 × (-26) – 5 × (-13)]

|A| = (-1/13)3 × 169 = -1/13

Cofactor of an element aij = Aij

A11 = (-1)1+1× (-1/13) = 1 × (-1/13) = -1/13

A12 = (-1)1+2× (-2/13) = (-1) × (-2/13) = 2/13

A13 = (-1)1+3× (-1/13) = 1 × (-1/13) = -1/13

A21 = (-1)2+1× (-2/13) = (-1) × (-2/13) = 2/13

A22 = (-1)2+2 × (3/13) = 1 × (-3/13) = -3/13

A23 = (-1)2+3 × (1/13) = (-1) × (1/13) = -1/13

A31 = (-1)3+1 × (-1/13) = 1 × (-1/13) = -1/13

A32 = (-1)3+2 × (1/13) = (-1) × 1/13 = -1/13

A33 = (-1)3+3 × (-5/13) = 1 × (-5/13) = -5/13

Adj (A-1) = = (A-1)-1 = Adj(A-1)/|A-1|

(A-1)-1 = = = A

(A-1)-1 = A

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