91Ó°ÊÓ

In matrix algebra, the determinant is a special number that can be calculated from a square matrix. For a 3x3 matrix like \( D(a, b, c) \), the determinant provides useful information about the matrix such as whether it is invertible or not. A key characteristic of matrices in the special linear group \( SL(3, \mathbb{R}) \) is that their determinants must always equal 1.

This implies that the matrices in this group are not only non-degenerate—which means they are invertible—but also preserve volume after transformation. To verify this for the given set \( H \), we compute the determinant of a generic matrix \( D(a, b, c) \). This computation was shown in the step-by-step solution to result in a determinant of 1: \[\det(D(a, b, c)) = \begin{vmatrix} 1 & a & b \0 & 1 & c \0 & 0 & 1 \end{vmatrix} = 1.\]By proving that every matrix in \( H \) maintains a determinant of 1, we confirm their inclusion in the special linear group.

When verifying if a particular set of matrices forms a subgroup of a given group like \( SL(3, \mathbb{R}) \), there are specific criteria to check.

These include:

• Identity: The identity matrix must be part of the subgroup.
• Closure: The subgroup must be closed under multiplication, meaning the product of any two matrices in the subgroup must also be in the subgroup.
• Inverses: Each matrix in the subgroup must have an inverse that is also in the subgroup.

For \( H = \{ D(a, b, c) \mid a, b, c \in \mathbb{R} \} \), we first confirmed that the identity matrix, \( D(0, 0, 0) \), is indeed part of \( H \). Afterward, both closure under multiplication and the presence of invertible elements were verified in the step-by-step solution using specific calculations. Since all these conditions are met, \( H \) is safely a subgroup.

The concept of inverse matrices is pivotal in linear algebra. An inverse matrix \( D^{-1} \) is defined such that when it is multiplied by the original matrix \( D \), the result is the identity matrix. For the matrices in \( H \), we need to confirm that not only do inverses exist but they also produce another matrix of the same form.

Given a matrix \( D(a, b, c) \), its inverse is:\[D(a, b, c)^{-1} = \begin{bmatrix} 1 & -a & ac-b \0 & 1 & -c \0 & 0 & 1 \end{bmatrix}.\]This inverse matrix satisfies the criteria of having all elements as real numbers and confirming a determinant of 1.

Thus, each matrix in \( H \) possesses an inverse that is also a part of \( H \), further establishing \( H \) as a subgroup.

Closure under multiplication is a critical property when determining if a set is a subgroup. It means if two matrices from a set are multiplied together, the resultant matrix should also belong to that set.

To ensure that \( H = \{ D(a, b, c) \mid a, b, c \in \mathbb{R} \} \) is closed under multiplication, consider two matrices, \( D(a_1, b_1, c_1) \) and \( D(a_2, b_2, c_2) \), within \( H \).

Their product is calculated as:\[D(a_1, b_1, c_1) \cdot D(a_2, b_2, c_2) = \begin{bmatrix} 1 & a_1 + a_2 & b_1 + a_2 c_1 + b_2 \0 & 1 & c_1 + c_2 \0 & 0 & 1 \end{bmatrix}.\]As shown in the solution, since the resulting matrix is still of the form \( D(a+b, b+ac+d, c+d) \), it is confirmed that any product of two matrices in \( H \) will remain within \( H \). Thus, the closure property is satisfied.