Question

Exercise \(\mathbf{3 . 2 . 3}\). Let \(A, B,\) and \(C\) denote \(n \times n\) matrices and assume that \(\operatorname{det} A=-1,\) det \(B=2,\) and \(\operatorname{det} C=3\). Evaluate: a. \(\operatorname{det}\left(A^{3} B C^{T} B^{-1}\right)\) b. \(\operatorname{det}\left(B^{2} C^{-1} A B^{-1} C^{T}\right)\)

Short answer

a. \(-3\); b. \(-2\)

Step-by-step solution

Understanding the Determinant of a Product

For part (a), consider the expression \( \operatorname{det}\left(A^{3} B C^{T} B^{-1}\right) \). The determinant of a product of matrices is the product of their determinants: \( \operatorname{det}(A^{3}) \times \operatorname{det}(B) \times \operatorname{det}(C^{T}) \times \operatorname{det}(B^{-1}). \)

Calculating Individual Determinants

Calculate \( \operatorname{det}(A^{3}) = (\operatorname{det}(A))^{3} = (-1)^{3} = -1 \). For \( \operatorname{det}(C^{T}) \), note that the determinant of a transpose is the same as the original matrix: \( \operatorname{det}(C^{T}) = \operatorname{det}(C) = 3 \). Finally, \( \operatorname{det}(B^{-1}) = \frac{1}{\operatorname{det}(B)} = \frac{1}{2} \).

Computing the Determinant of Part a

Plug these values into the formula from Step 1: \( \operatorname{det}(A^{3}) \times \operatorname{det}(B) \times \operatorname{det}(C^{T}) \times \operatorname{det}(B^{-1}) = (-1) \times 2 \times 3 \times \frac{1}{2} = -3 \).

Understanding the Determinant of Another Product

For part (b), consider the expression \( \operatorname{det}(B^{2} C^{-1} A B^{-1} C^{T}) \). Again, the determinant of a product of matrices is the product of their determinants.

Calculating Individual Determinants for Part b

Calculate \( \operatorname{det}(B^{2}) = (\operatorname{det}(B))^{2} = 2^{2} = 4 \). For \( \operatorname{det}(C^{-1}) \), use \( \frac{1}{\operatorname{det}(C)} = \frac{1}{3} \). Already calculated values: \( \operatorname{det}(A) = -1 \) and \( \operatorname{det}(B^{-1}) = \frac{1}{2} \). Also, \( \operatorname{det}(C^{T}) = 3 \).

Computing the Determinant of Part b

Plug these values into the formula: \( \operatorname{det}(B^{2}) \times \operatorname{det}(C^{-1}) \times \operatorname{det}(A) \times \operatorname{det}(B^{-1}) \times \operatorname{det}(C^{T}) = 4 \times \frac{1}{3} \times -1 \times \frac{1}{2} \times 3 = -2 \).

Key concepts

Matrix Transpose

When we transpose a matrix, we essentially "flip" it across its diagonal. This means that each element \(a_{ij}\) in the original matrix \(A\) becomes \(a_{ji}\) in the transposed matrix \(A^T\). So, the rows transform into columns and vice versa.
It's important to note that the size or dimensions of the matrix do not change when transposing. If you have a 3x3 matrix, its transpose is also a 3x3 matrix.

• The determinant of a transposed matrix remains the same as the original matrix. This is why in our exercise, \(\operatorname{det}(C^T) = \operatorname{det}(C)\).
• Transposing twice will lead you back to the original matrix, i.e., \(\left(A^T\right)^T = A\).
• Additionally, transposition is distributive over addition: \(\left(A + B\right)^T = A^T + B^T\), and reverses the order of a product: \(\left(A \cdot B\right)^T = B^T \cdot A^T\).

Inverse Matrix

An inverse matrix is like the 'opposite' of a given matrix. If you multiply a matrix by its inverse, you get the identity matrix, much like multiplying a number by its reciprocal results in 1.
For a square matrix \(A\), its inverse \(A^{-1}\) satisfies the equation \(A \cdot A^{-1} = I \), where \(I\) is the identity matrix. If a matrix doesn't have an inverse, it's called singular or non-invertible.

• Calculating the inverse of a matrix is not always straightforward and requires that the determinant of the matrix is not zero. Only non-zero determinant matrices can have an inverse.
• The inverse of a product of matrices can be determined by taking the inverse of each individual matrix, in reverse order: \(\left(AB\right)^{-1} = B^{-1}A^{-1}\).
• In our exercise, we used the determinant of \(B^{-1}\), which is \(\frac{1}{\operatorname{det}(B)}\), to solve the problems.

Determinant Properties

Determinants are a unique number associated with square matrices. They hold important properties that aid in various calculations, especially in linear algebra.
One crucial property is that the determinant of a matrix product equals the product of their determinants. Hence, \(\operatorname{det}(A \cdot B) = \operatorname{det}(A) \cdot \operatorname{det}(B)\).

• In our exercise, determinant properties were vital in simplifying products of matrices for quick computation.
• Additionally, raising a matrix to a power raises its determinant to that power: \(\operatorname{det}(A^k) = \left(\operatorname{det}(A)\right)^k\).
• For the inverse matrix, its determinant is the reciprocal: \(\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}\) if \(A\) is invertible.

These properties simplify complex matrix expressions down to simpler numeric calculations as demonstrated in our evaluation problems.

Matrix Multiplication

Matrix multiplication is different from regular number multiplication. To multiply two matrices \(A\) and \(B\), the number of columns in \(A\) must match the number of rows in \(B\). Each element of the resulting matrix is computed as the dot product of rows and columns.
One should always pay attention to the order of multiplication, as matrix multiplication is not commutative. This means \(A \cdot B eq B \cdot A\) in general.

• Associativity holds in matrix multiplication, allowing us to group matrices: \(A \cdot (B \cdot C) = (A \cdot B) \cdot C\).
• In our exercise, since matrix multiplication affected the determinant, we utilized the associative property to simplify calculations by reordering matrix products strategically.

Think of matrix multiplication like combining transformations or rotations. Each multiplication results in a new transformation applied on a given data set or shape.