Question

Let \(A\) and \(B\) be invertible \(n \times n\) matrices. Evaluate: a. \(\operatorname{det}\left(B^{-1} A B\right)\) b. \(\operatorname{det}\left(A^{-1} B^{-1} A B\right)\)

Short answer

a. \(\operatorname{det}(A)\), b. 1.

Step-by-step solution

Applying Properties of Determinants

We know that for any square matrix \(X\), \(\operatorname{det}(X^{-1}) = \frac{1}{\operatorname{det}(X)}\). Also, the determinant of a product of matrices is the product of their determinants, i.e., \(\operatorname{det}(XY) = \operatorname{det}(X)\cdot\operatorname{det}(Y)\). We'll use these properties to solve the problems.

Evaluate \(\operatorname{det}(B^{-1}AB)\)

Using the determinant properties, we have:\[\operatorname{det}(B^{-1}AB) = \operatorname{det}(B^{-1}) \cdot \operatorname{det}(A) \cdot \operatorname{det}(B)\]Since \(B\) is invertible, \(\operatorname{det}(B^{-1}) = \frac{1}{\operatorname{det}(B)}\). Therefore:\[\operatorname{det}(B^{-1}AB) = \frac{1}{\operatorname{det}(B)} \cdot \operatorname{det}(A) \cdot \operatorname{det}(B)\]This simplifies to:\[\operatorname{det}(B^{-1}AB) = \operatorname{det}(A)\]

Evaluate \(\operatorname{det}(A^{-1} B^{-1} A B)\)

Again, using the determinant properties:\[\operatorname{det}(A^{-1} B^{-1} A B) = \operatorname{det}(A^{-1}) \cdot \operatorname{det}(B^{-1}) \cdot \operatorname{det}(A) \cdot \operatorname{det}(B)\]We know \(\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}\) and \(\operatorname{det}(B^{-1}) = \frac{1}{\operatorname{det}(B)}\), thus:\[\operatorname{det}(A^{-1} B^{-1} A B) = \frac{1}{\operatorname{det}(A)} \cdot \frac{1}{\operatorname{det}(B)} \cdot \operatorname{det}(A) \cdot \operatorname{det}(B)\]This simplifies to:\[\operatorname{det}(A^{-1} B^{-1} A B) = 1\]

Key concepts

Understanding Invertible Matrices

In linear algebra, invertible matrices play a crucial role. When we say a matrix is invertible, it means that there is another matrix which can reverse the effect of the original matrix via multiplication. If a matrix is represented as \(A\), then its inverse is denoted as \(A^{-1}\). A key point to remember is:

• The inverse of a matrix \(A\) is defined such that \(A \cdot A^{-1} = I\), where \(I\) is the identity matrix. The identity matrix is like the number 1 for scalars, maintaining other elements in a multiplication unaffected.
• Only square matrices (same number of rows and columns) can potentially be invertible. However, not all square matrices are invertible. A matrix must have a non-zero determinant to be invertible.

If you need to check if a given \(n \times n\) matrix is invertible, simply calculate its determinant. If \(\operatorname{det}(A) eq 0\), then the matrix \(A\) is invertible. Otherwise, it isn't. This makes determinants a vital tool in determining invertibility.

The Essence of Matrix Multiplication

Matrix multiplication is fundamental in linear algebra, serving essential operations in various disciplines. The key idea is that you can "combine" matrices in such a way that represents subsequent transformations or operations.

• To multiply two matrices, say \(A\) and \(B\), the number of columns in \(A\) must match the number of rows in \(B\). This results in a new matrix that represents a combined transformation.
• The multiplication operation is not commutative; generally, \(A \cdot B eq B \cdot A\). However, associative property holds, meaning \((AB)C = A(BC)\).

Matrix multiplication can be visualized as applying transformation \(B\) first, followed by transformation \(A\). Therefore, matrix multiplication allows us to bring multiple operations together. This concept is crucial when simplifying expressions involving determinants, as it helps to manipulate and rearrange matrices effectively.

Exploring Properties of Determinants

Determinants reveal a lot about a matrix, including whether it's invertible and how transformations like rotations, scaling, or shears affect space.

• The determinant of a matrix product is the product of the determinants of the individual matrices: \(\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B)\). This property is pivotal in solving the exercise above.
• The determinant of an inverse matrix is the reciprocal of the determinant of the original matrix: \(\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}\), assuming \(A\) is invertible.
• A matrix with zero determinant is singular, indicating it's non-invertible.

These determinant properties were used to deduce that for \(\operatorname{det}(B^{-1}AB)\), it simplifies to \(\operatorname{det}(A)\). Additionally, they help us understand why the determinant of \(A^{-1}B^{-1}AB\) simplifies to 1, employing the fact that multiplicative inverses negate each other's effects when combined sequentially.