Given: A is a square matrix of order 3, such that A(adj A) = 10I

To find: |adj A|

Solution:

1. Using the property of adjoint of a matrix

We know that, adj(A) = (Cofactor of A)T

where Cofactor of A is the matrix obtained by taking the determinant of each minor of A and multiplying it by (-1)^(i+j), where i and j are the row and column indices of the element.

So, A(adj A) = A((Cofactor of A)T) = (ACofactor of A)T

2. Using the given condition

We are given that A(adj A) = 10I

Substituting this in the above equation, we get:

(ACofactor of A)T = 10I

Taking determinant on both sides, we get:

|ACofactor of A|T = 10^3

|ACofactor of A| = 10^3 (since determinant of a matrix is equal to the determinant of its transpose)

3. Using the property of determinant

We know that, |AB| = |A||B|

Substituting A = adj(A), we get:

|adj(A)Cofactor of A| = |adj(A)||Cofactor of A|

Since adj(A)Cofactor of A = |A|I (where I is the identity matrix), we get:

|A||Cofactor of A| = |adj(A)||Cofactor of A|

|A| = |adj(A)|

Substituting this in the previous equation, we get:

|adj(A)| |Cofactor of A| = 10^3

|adj(A)| = (10^3)/|Cofactor of A|

4. Finding the value of |Cofactor of A|

Since A is a square matrix of order 3, its adjoint matrix adj(A) is of order 3. Therefore, the Cofactor of A will be a matrix of order 3 as well.

Using the formula for finding the Cofactor of a matrix, we get:

Cofactor of A = (−1)^{i+j} M_{ij}

where M_{ij} is the determinant of the matrix obtained by deleting the i-th row and j-th column of A.

So, we need to find the determinant of 9 matrices (3x3) to find the Cofactor of A. However, we can simplify this process by using the property of symmetry of the Cofactor matrix.

We know that the Cofactor matrix is symmetric, i.e., Cofactor of A = (Cofactor of A)T

Therefore, we can find the determinant of only 4 matrices and use them to find the determinant of the remaining 5 matrices.

The 4 matrices are:

M_{11} = det\begin{pmatrix}a_{22} & a_{23}\\a_{32} & a_{33}\end{pmatrix}

M_{22} = det\begin{pmatrix}a_{11} & a_{13}\\a_{31} & a_{33}\end{pmatrix}

M_{33} = det\begin{pmatrix}a_{11} & a_{12}\\a_{21} &