Find the value of x, y and z, if A =satisfies A’ = A¹

OR

Verify: A(adjA) = (adjA)A = |A|I for matrix A =

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 aij} 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 aij} of given by A_ij = (– 1)^i+j M_ij

_And

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

_{|A| = a11}A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

_And

If then,

where, A_ij is cofactor of a_ij

Calculating for

We get,

a₁₁ = 1, A₁₁ = 0

a₁₂ = -1, A₁₂ = -11

a₁₃ = 2, A₁₃ = 0

a₂₁ = 3, A₂₁ = 3

a₂₂ = 0, A₂₂ = 1

a₂₃ = -2, A₂₃ = -1

a₃₁ = 1, A₃₁ = 2

a₃₂ = 3, A₃₂ = 8

a₃₃ = 3, A₃₃ = 3

∴ |A| = 1(0) + (-1) (-11) + 2(0) = 11

Now,

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