Problem 39 - Differential Equations and Linear Algebra

Question

Let \(A\) and \(B\) be \(4 \times 4\) matrices such that \(\operatorname{det}(A)=5\) and \(\operatorname{det}(B)=3 .\) Compute the determinant of the given matrix. $$\left(A^{-1} B^{2}\right)^{3}$$

Step-by-Step Solution

Answer

The determinant of the matrix \(\left(A^{-1} B^{2}\right)^{3}\) is \(\boxed{\frac{729}{125}}\).

Step 1: Determine the determinants of \(A^{-1}\) and \(B^{2}\)

We are given that \(\operatorname{det}(A) = 5\) and \(\operatorname{det}(B) = 3\). We will now find the determinants of \(A^{-1}\) and \(B^2\). 1.1 For \(A^{-1}\), we know that: \[\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}\] Thus, \[\operatorname{det}(A^{-1}) = \frac{1}{5}\] 1.2 For \(B^{2}\), we use the property that: \[\operatorname{det}(B^{2}) = (\operatorname{det}(B))^2\] So, \[\operatorname{det}(B^{2}) = 3^2 = 9\]

Step 2: Compute the determinant of the matrix product \(A^{-1} B^{2}\)

With the determinants of \(A^{-1}\) and \(B^2\) calculated, we can now compute the determinant of their product, \(A^{-1} B^{2}\), using the property that: \[\operatorname{det}(A^{-1} B^{2}) = \operatorname{det}(A^{-1}) \cdot \operatorname{det}(B^{2})\] Therefore, \[\operatorname{det}(A^{-1} B^{2}) = \frac{1}{5} \cdot 9 = \frac{9}{5}\]

Step 3: Compute the determinant of \((A^{-1} B^{2})^3\)

Finally, we need to compute the determinant of the given matrix \((A^{-1} B^{2})^{3}\). Using the property that: \[\operatorname{det}((A^{-1} B^{2})^{3}) = (\operatorname{det}(A^{-1} B^{2}))^3\] Calculate the value: \[\operatorname{det}((A^{-1} B^{2})^{3}) = \left(\frac{9}{5}\right)^3 = \frac{729}{125}\] The determinant of the matrix \(\left(A^{-1} B^{2}\right)^{3}\) is \(\boxed{\frac{729}{125}}\).

Key Concepts

Determinant Properties

Understanding determinant properties is essential for solving many linear algebra problems. The determinant of a matrix, denoted as \( \operatorname{det}(A) \), is a scalar value that serves as a powerful tool in matrix theory, providing insight into matrix characteristics.

Key properties of determinants include:

• Multiplicative: For any two square matrices \( A \) and \( B \), \( \operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B) \). This property was crucial in the exercise to find the determinant of \( A^{-1} B^{2} \).
• Inversion: The determinant of an inverse matrix \( A^{-1} \) is the reciprocal of the determinant of the matrix \( A \), that is, \( \operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)} \).
• Exponents: For a matrix \( A \) raised to a power \( n \), \( \operatorname{det}(A^n) = (\operatorname{det}(A))^n \). This property assisted in calculating \( \operatorname{det}(B^{2}) \) and \( \operatorname{det}((A^{-1} B^{2})^{3}) \) in the given problem.

These properties work together to simplify complex matrix operations, allowing us to find determinants of products and exponents of matrices with relative ease.

Matrix Inversion

In matrix theory, finding the inverse of a matrix \( A \)—denoted as \( A^{-1} \)—is a key operation that can be thought of as a division analogue in the realm of matrices. An inverse matrix, when multiplied by its original matrix, yields the identity matrix \( I \), symbolically shown as \( AA^{-1} = A^{-1}A = I \).

However, not all matrices have inverses. A matrix must be square and have a non-zero determinant to have an inverse. This implies that the matrix is 'invertible' and the system of equations it represents has a unique solution. In the exercise, the property of the determinant of an inverse matrix (\( \operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)} \) ) was applied to find \( \operatorname{det}(A^{-1}) \) given that \( \operatorname{det}(A) = 5 \).

Matrix Multiplication

Matrix multiplication is not commutative like standard number multiplication—meaning \( AB \) may not equal \( BA \). For matrix multiplication to be valid, the number of columns in the first matrix must equal the number of rows in the second matrix.

Additionally, matrix multiplication is distributive over matrix addition, associative, and has an identity element, which is the identity matrix for the given dimension. In terms of determinants, it is key to remember that the determinant of a product is equal to the product of the determinants (\( \operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B) \)). This concept was used in the given exercise to compute \( \operatorname{det}(A^{-1} B^{2}) \), simplifying the process significantly.

Exponent of Matrices

Raising a matrix to a power involves multiplying the matrix by itself a number of times. The exponent must be a non-negative integer, and this is only possible with square matrices. One important determinant property related to matrix exponents states that \( \operatorname{det}(A^n) = (\operatorname{det}(A))^n \), which was employed in the exercise to find the determinant of \( (A^{-1} B^{2})^{3} \).

This powerful property allows us to avoid the cumbersome task of actual matrix multiplication when only the determinant is needed. Instead, we can calculate the determinant once and then raise it to the power, as we can see with \( \operatorname{det}(B^{2}) = (\operatorname{det}(B))^2 \) and \( \operatorname{det}((A^{-1} B^{2})^{3}) = (\operatorname{det}(A^{-1} B^{2}))^3 \) from the solution steps. This exponent rule simplifies what would otherwise be a highly labor-intensive process.