let A be a nonsingular square matrix of order 3 multiply 3 then find [adjA]

To find the adjugate of a nonsingular square matrix \( A \) of order \( 3 \times 3 \), we first need to understand what the adjugate (or adjoint) of a matrix is. The adjugate of a matrix is the transpose of its cofactor matrix. Let's break this down step by step.

Understanding the Matrix and Its Cofactors

Given a matrix \( A \) of order \( 3 \times 3 \), it can be represented as:

\( A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \)

To find the adjugate, we first need to compute the cofactor matrix. The cofactor \( C_{ij} \) of an element \( a_{ij} \) is calculated as:

• Calculate the determinant of the \( 2 \times 2 \) matrix that remains after removing the \( i \)-th row and \( j \)-th column from \( A \).
• Multiply this determinant by \( (-1)^{i+j} \) to account for the position of the element.

Calculating the Cofactor Matrix

For our \( 3 \times 3 \) matrix \( A \), the cofactor matrix \( C \) will be:

\( C = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix} \)

Where each \( C_{ij} \) is calculated as follows:

• \( C_{11} = (-1)^{1+1} \cdot \text{det} \begin{pmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \)
• \( C_{12} = (-1)^{1+2} \cdot \text{det} \begin{pmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} \)
• \( C_{13} = (-1)^{1+3} \cdot \text{det} \begin{pmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} \)
• And similarly for \( C_{21}, C_{22}, C_{23}, C_{31}, C_{32}, C_{33} \).

Finding the Adjugate

Once we have the cofactor matrix \( C \), the adjugate \( \text{adj}(A) \) is simply the transpose of this cofactor matrix:

\( \text{adj}(A) = C^T \)

This means that the rows of the cofactor matrix become the columns of the adjugate matrix. For example, if:

\( C = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix} \)

Then:

\( \text{adj}(A) = \begin{pmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{pmatrix} \)

Example Calculation

Let’s consider a specific example to illustrate this process. Suppose:

\( A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \)

We can calculate the cofactors:

• \( C_{11} = \text{det} \begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} = (1)(0) - (4)(6) = -24 \)
• \( C_{12} = -\text{det} \begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = -(-20) = 20 \)
• \( C_{13} = \text{det} \begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} = (0)(6) - (1)(5) = -5 \)
• Continuing this process for \( C_{21}, C_{22}, C_{23}, C_{31}, C_{32}, C_{33} \).

After calculating all cofactors, you would then transpose the cofactor matrix to find the adjugate.

Final Thoughts

The adjugate matrix is particularly useful in finding the inverse of a matrix, as it relates to the formula:

\( A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \)

Thus, understanding how to compute the adjugate is a foundational skill in linear algebra, especially when dealing with matrix inverses and determinants.