Answer :

A’ (Transpose of A) is defined by

If A = [a_{ij}]_{m × n} then A’ = [a_{ji}]_{n × m}

Therefore, we have

A’ = A

Comparing corresponding elements, we have

x = 2y

x = z

y = z

Therefore, general solution will be

x = 2z, y = z, z = z

[Infinitely many solutions possible]

**OR**

_{We know,}

_{Minor}_{of an element a}_{ij} of the determinant of matrix A is the determinant obtained by deleting ith row and jth column and denoted by M_{ij}

_{and}

_{Cofactor}_{of a}_{ij} of given by A_{ij} = (– 1)^{i+j} M_{ij}

_{And}

_{Value of determinant}_{of a matrix A is obtained by}

_{|A| = a}_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}

_{And}

If then,

where, A_{ij} is cofactor of a_{ij}

Calculating for

We get,

a_{11} = 1, A_{11} = 0

a_{12} = -1, A_{12} = -11

a_{13} = 2, A_{13} = 0

a_{21} = 3, A_{21} = 3

a_{22} = 0, A_{22} = 1

a_{23} = -2, A_{23} = -1

a_{31} = 1, A_{31} = 2

a_{32} = 3, A_{32} = 8

a_{33} = 3, A_{33} = 3

∴ |A| = 1(0) + (-1) (-11) + 2(0) = 11

Now,

∴ it is verified that, A(adjA) = (adjA)A = |A|I

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