Problem 7 - Advanced Engineering Mathematics

Question

In Problems 1-20, fill in the blanks or answer true/false. If \(\mathbf{A}\) is a \(3 \times 3\) matrix such that \(\operatorname{det} \mathbf{A}=5\), then \(\operatorname{det}\left(\frac{1}{2} \mathbf{A}\right)=\) _________ and \(\operatorname{det}\left(-\mathbf{A}^{T}\right)=\) _________

Step-by-Step Solution

Answer

\( \operatorname{det}\left(\frac{1}{2} \mathbf{A}\right) = \frac{5}{8} \) and \( \operatorname{det}(-\mathbf{A}^T) = -5 \).

Step 1: Understanding Determinant Scaling

Given that \( \operatorname{det}(\mathbf{A}) = 5 \) for a \( 3 \times 3 \) matrix, applying a scalar \( \frac{1}{2} \) to the matrix \( \mathbf{A} \) scales the determinant by the cube of the scalar: \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \). Hence, \( \operatorname{det}\left(\frac{1}{2} \mathbf{A}\right) = \frac{1}{8} \times 5 = \frac{5}{8} \).

Step 2: Determinant Property with Transpose and Negative Sign

For the determinant of \(-\mathbf{A}^T\), note that the determinant of a transpose matrix is the same as the original matrix: \( \operatorname{det}(\mathbf{A}^T) = \operatorname{det}(\mathbf{A}) = 5 \). Multiplying the matrix by \(-1\), flips the sign of the determinant, hence \( \operatorname{det}(-\mathbf{A}^T) = -1 \times 5 = -5 \).

Key Concepts

Matrix Scaling

When you scale a matrix by multiplying it by a scalar, this affects the matrix's determinant in a predictable way. For a square matrix, instead of just multiplying the determinant by the scalar, you multiply it by the scalar raised to the power of the matrix's size. Imagine you have a matrix \( \mathbf{A} \) of size \( 3 \times 3 \), and you scale it by \( \frac{1}{2} \).
This means you multiply the determinant by \( \left(\frac{1}{2}\right)^3 \), which equals \( \frac{1}{8} \), because for a \( 3 \times 3 \) matrix, the power is 3. Therefore, if \( \operatorname{det}(\mathbf{A}) = 5 \), then \( \operatorname{det}\left(\frac{1}{2} \mathbf{A}\right) = \frac{5}{8} \).
It's crucial to remember:

• The determinant of a scaled matrix depends on the size of the matrix.
• You raise the scalar to the power of the matrix's dimension.
• This applies to square matrices only.

Using these rules helps in determining how changing the size or scale of elements in a matrix impacts its determinant.

Transpose of a Matrix

The transpose of a matrix \( \mathbf{A} \), denoted as \( \mathbf{A}^{T} \), is a new matrix created by swapping the rows and columns of \( \mathbf{A} \). This process might seem complex, but the properties of the determinant make it straightforward. Interestingly, the determinant doesn't change its value when a matrix is transposed.
This means:

• If \( \operatorname{det}(\mathbf{A}) = 5 \), then \( \operatorname{det}(\mathbf{A}^T) \) is also 5.
• The orientation of data inside the matrix changes, but its determinant remains invariant.
• The transpose operation is particularly useful in solving systems of equations and simplifying expressions.

In the context of the exercise, knowing that the determinant remains unaffected by transposition helps simplify the calculation of \( -\mathbf{A}^{T}\), allowing us to focus solely on the impact of negative scaling.

Determinant Properties

The determinant is a powerful tool in linear algebra that provides insights into matrix behavior. Its properties come in handy for matrix manipulations.Here are some key properties regarding how determinants behave:

• The determinant of a product of matrices equals the product of the determinants: \( \operatorname{det}(\mathbf{AB}) = \operatorname{det}(\mathbf{A}) \cdot \operatorname{det}(\mathbf{B}) \).
• For scalar multiplication, as seen in the exercise, multiplying a matrix by a scalar affects its determinant, raising the scalar by the power of the matrix's size.
• If \( \mathbf{A} \) is a \( n \times n \) matrix, then \( \operatorname{det}(-\mathbf{A}) = (-1)^n \operatorname{det}(\mathbf{A}) \), meaning it changes the sign if \( n \) is odd.
• The determinant of a transpose matrix remains the same \( \operatorname{det}(\mathbf{A}^T) = \operatorname{det}(\mathbf{A}) \).

In our problem, these properties simplify calculations significantly. Applying them helps us understand why \( \operatorname{det}(-\mathbf{A}^T) = -5 \) when \( \operatorname{det}(\mathbf{A}) = 5 \), since multiplying by a negative one, given our 3x3 matrix, flips the determinant's sign.