This solution explains how to find the determinant of \(\det(B^2 A^{-1} (B^T)^3)\) given information about the determinants of other matrix expressions involving invertible matrices \(A\) and \(B\).

Understanding Determinant Properties for Invertible Matrices

To solve this problem, we need to use some fundamental properties of determinants, especially for invertible matrices like \(A\) and \(B\):

• Product Rule: The determinant of a product of matrices is the product of their determinants. For matrices \(X\) and \(Y\), \(\det(XY) = \det(X) \det(Y)\).
• Power Rule: The determinant of a matrix raised to a power \(n\) is the determinant raised to that power. \(\det(X^n) = (\det(X))^n\).
• Transpose Rule: The determinant of the transpose of a matrix is the same as the determinant of the original matrix. \(\det(X^T) = \det(X)\).
• Inverse Rule: The determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix. \(\det(X^{-1}) = \frac{1}{\det(X)}\).

These properties hold true for square matrices, and since \(A\) and \(B\) are invertible matrices of order \(3 \times 3\), we can apply them.

Analyzing the Given Determinant Information

We are given two equations involving the determinants of matrix expressions:

1. \(\det(A^2 B (A^T)^3) = 16\)
2. \(\det(A^3 B^{-2}) = 32\)

Let's simplify these using the determinant properties. Let \(\det(A) = x\) and \(\det(B) = y\).

Simplifying the First Equation

Using the Product Rule, Power Rule, and Transpose Rule:

\(\det(A^2 B (A^T)^3) = \det(A^2) \det(B) \det((A^T)^3)\)

Applying the Power Rule:

\(\det(A^2) = (\det(A))^2 = x^2\) \(\det((A^T)^3) = (\det(A^T))^3 = (\det(A))^3 = x^3\)

So, the first equation becomes:

\((x^2) \cdot y \cdot (x^3) = 16\) \(x^5 y = 16\) (Equation 1)

Simplifying the Second Equation

Using the Power Rule and Inverse Rule:

\(\det(A^3 B^{-2}) = \det(A^3) \det(B^{-2})\)

Applying the Power Rule and Inverse Rule:

\(\det(A^3) = (\det(A))^3 = x^3\) \(\det(B^{-2}) = \det((B^{-1})^2) = (\det(B^{-1}))^2 = \left(\frac{1}{\det(B)}\right)^2 = \frac{1}{y^2}\)

So, the second equation becomes:

\(x^3 \cdot \frac{1}{y^2} = 32\) \(\frac{x^3}{y^2} = 32\) (Equation 2)

Calculating Determinants of A and B

Now we have a system of two equations with two variables, \(\det(A) = x\) and \(\det(B) = y\):

1. \(x^5 y = 16\)
2. \(\frac{x^3}{y^2} = 32\)

We can solve this system. From Equation 1, we can express \(y\) in terms of \(x\):

\(y = \frac{16}{x^5}\)

Now, substitute this expression for \(y\) into Equation 2:

\(\frac{x^3}{\left(\frac{16}{x^5}\right)^2} = 32\) \(\frac{x^3}{\frac{16^2}{(x^5)^2}} = 32\) \(\frac{x^3}{\frac{256}{x^{10}}} = 32\) \(\frac{x^3 \cdot x^{10}}{256} = 32\) \(\frac{x^{13}}{256} = 32\)

Multiply both sides by 256:

\(x^{13} = 32 \times 256\)

Expressing the numbers as powers of 2:

\(32 = 2^5\) \(256 = 2^8\) \(x^{13} = 2^5 \times 2^8 = 2^{5+8} = 2^{13}\)

Therefore, \(\det(A) = x = 2\).

Now substitute the value of \(x\) back into the expression for \(y\):

\(y = \frac{16}{x^5} = \frac{16}{2^5} = \frac{16}{32} = \frac{1}{2}\)

So, \(\det(B) = y = \frac{1}{2}\).

Determining the Target Determinant Value

We need to find the value of \(\det(B^2 A^{-1} (B^T)^3)\).

Let's apply the determinant properties again:

\(\det(B^2 A^{-1} (B^T)^3) = \det(B^2) \det(A^{-1}) \det((B^T)^3)\)

Using the Power Rule, Inverse Rule, and Transpose Rule:

\(\det(B^2) = (\det(B))^2 = y^2\) \(\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{x}\) \(\det((B^T)^3) = (\det(B^T))^3 = (\det(B))^3 = y^3\)

Substituting these back into the expression:

\(\det(B^2 A^{-1} (B^T)^3) = y^2 \cdot \frac{1}{x} \cdot y^3 = \frac{y^5}{x}\)

Substituting Calculated Values

We found that \(\det(A) = x = 2\) and \(\det(B) = y = \frac{1}{2}\).

Substitute these values into the expression \(\frac{y^5}{x}\):

\(\det(B^2 A^{-1} (B^T)^3) = \frac{\left(\frac{1}{2}\right)^5}{2}\) \(= \frac{\frac{1}{32}}{2}\) \(= \frac{1}{32 \times 2}\) \(= \frac{1}{64}\)

Final Result

The value of \(\det(B^2 A^{-1} (B^T)^3)\) is \(\frac{1}{64}\).