Determinant Calculation for 3x3 Matrices Using Multiplicativity Property

When working with matrices, the properties of determinants are often used to simplify calculations. In this article, we will delve into a specific problem involving the determinant of a 3x3 matrix (A), given that (AB 2I) and (B 6). We will utilize the property of determinants, particularly their (text{multiplicativity}), to find (det(A)).

Understanding the Problem

The given matrix equation is

(AB 2I)

where (A) and (B) are 3x3 matrices, and (I) is the identity matrix. Additionally, it is given that (B 6).

Using the Multiplicativity Property of Determinants

The property of determinants that we will use is their (text{multiplicativity}). For any two square matrices (A) and (B), the determinant of their product is the product of their determinants:

(det(AB) det(A) cdot det(B))

Given that (AB 2I), we can apply this property to get:

(det(AB) det(2I))

Since (2I) represents a scalar multiple of the identity matrix (I), the determinant of (2I) can be determined as:

(det(2I) 2^3 8)

Solving for (det(A))

Substituting these values into the multiplicativity property, we get:

(det(A) cdot det(B) 8)

Given (det(B) 6), we can solve for (det(A)):

(det(A) cdot 6 8)

(det(A) frac{8}{6} frac{4}{3})

Thus, the determinant of matrix (A) is (frac{4}{3}).

Technical Note

A simpler and more usual notation for determinants is often used, such as (det(A)). This notation emphasizes that the entire matrix is being evaluated.

Alternative Approach and Potential Weakness

Another way to solve this problem involves constructing matrix (B) explicitly. If we assume (B 6I), where (I) is the identity matrix, then:

(det(B) 6)

Let (A) be a matrix such that:

(A begin{bmatrix} frac{1}{3} 0 0 0 1 0 0 0 1 end{bmatrix})

Then, the product (AB) can be verified to equal (2I) (or (2I_3)). In this case, the transpose of (A) is equal to (A) itself, i.e., (A^T A).

The determinant of the transpose of (A) is the same as the determinant of (A) itself:

(det(A^T) det(A) 2/6 1/3)

Conclusion

In summary, using the multiplicativity property of determinants, we find that

(det(A) frac{4}{3})

This method is robust and provides a straightforward solution to the problem, ensuring that the properties of determinants are correctly applied.