https://file.aiquickdraw.com/mathbot/file/1725237230821hsjmblos.jpg

To solve problem 7(i) from the provided assignment, let's first understand the problem statement and then proceed with the solution.

### Problem 7(i) Statement:
The problem asks us to prove the following identity:

\[
\det(A) = \sum_{i=1}^{n} a_{ij} A_{ij}
\]

where:

- \( A \) is an \( n \times n \) matrix.
- \( a_{ij} \) is the element in the \( i \)-th row and \( j \)-th column of the matrix \( A \).
- \( A_{ij} \) is the cofactor of the element \( a_{ij} \).

### Explanation and Proof:
This statement is a well-known result in linear algebra, known as the cofactor expansion or Laplace expansion of the determinant along the \( j \)-th column.

1. **Determinant Definition:**

   The determinant of an \( n \times n \) matrix \( A \) can be expanded along any row or column. Here, we are expanding along the \( j \)-th column.

2. **Cofactor Definition:**

   The cofactor \( A_{ij} \) of the element \( a_{ij} \) is defined as:

   \[
   A_{ij} = (-1)^{i+j} M_{ij}
   \]

   where \( M_{ij} \) is the determinant of the \((n-1) \times (n-1)\) matrix obtained by deleting the \( i \)-th row and \( j \)-th column from \( A \).

3. **Cofactor Expansion along the \( j \)-th Column:**

   The determinant of \( A \) is given by the sum of the products of the elements in any column (here, the \( j \)-th column) and their corresponding cofactors:

   \[
   \det(A) = \sum_{i=1}^{n} a_{ij} A_{ij}
   \]

   This formula essentially says that the determinant of \( A \) can be computed by taking each element in the \( j \)-th column, multiplying it by its cofactor, and summing these products.

4. **Proof Outline:**

   To prove this, consider the definition of the determinant for an \( n \times n \) matrix using cofactor expansion. Specifically, expanding along the \( j \)-th column involves summing the products of each element in the column by the determinant of the submatrix that remains after removing the row and column containing that element.

### Conclusion:
This identity is a direct application of the cofactor expansion method for computing determinants. Thus, the expression \( \det(A) = \sum_{i=1}^{n} a_{ij} A_{ij} \) holds true by the definition of the determinant.

---

Would you like any more details or have any further questions about this proof?

### Related Questions:
1. What is the significance of the sign \( (-1)^{i+j} \) in the cofactor formula?
2. How would the expansion look if we were expanding along a row instead of a column?
3. Can this identity be used to simplify the calculation of determinants for larger matrices?
4. How does the determinant help in determining whether a matrix is invertible?
5. Can the cofactor expansion be applied to non-square matrices?

### Tip:
When calculating determinants, always choose the row or column with the most zeros for expansion to simplify the calculation.

Learn how to prove the identity where the determinant of a matrix is expressed as the sum of products of elements and their corresponding cofactors along a specified column. This problem covers advanced linear algebra concepts, including determinants, cofactor expansion, and theorem application. Suitable for advanced undergraduate level students and enthusiasts of linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation