To solve the given problems for each matrix, we will follow these steps:
(i) Check if the matrix is in reduced row echelon form (RREF). If it is, we will note "yes". If not, we will compute the RREF of the matrix.
(ii) Calculate the determinant of the matrix. Note that the determinant is only defined for square matrices.
(iii) Solve the homogeneous system of linear equations \(\mathbf{A} \overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{0}}\) to find all solutions.
The matrix \(\begin{bmatrix} 1 & 3 & 4 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}\) is in reduced row echelon form because:
- The leading entry in each nonzero row is 1.
- Each leading 1 is the only nonzero entry in its column.
- The leading 1 in each row is to the right of the leading 1 in the row above it.
Thus, the answer is yes.
The matrix \(\begin{bmatrix} 1 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 1 & -2 \end{bmatrix}\) is not in reduced row echelon form because the leading 1 in the third row is not to the right of the leading 1 in the first row. The RREF of this matrix is:
\[
\begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}
\]
The matrix \(\begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}\) is in reduced row echelon form because it satisfies all the conditions for RREF.
Thus, the answer is yes.
The determinant of the matrix \(\begin{bmatrix} 1 & 3 & 4 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}\) is calculated as:
\[
\det = 1 \cdot (2 \cdot 1 - 0 \cdot 1) = 2
\]
The determinant of the matrix \(\begin{bmatrix} 1 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 1 & -2 \end{bmatrix}\) is 0 because it contains a row of zeros.
The determinant is not defined for the matrix \(\begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}\) because it is not a square matrix.
The matrix \(\begin{bmatrix} 1 & 3 & 4 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}\) has a unique solution \(\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{0}}\) because it is full rank.
The matrix \(\begin{bmatrix} 1 & 0 & 4 \\ 0 & 0 & 0 \\ 0 & 1 & -2 \end{bmatrix}\) has a solution set given by:
\[
\overrightarrow{\mathbf{x}} = t \begin{bmatrix} -4 \\ 2 \\ 1 \end{bmatrix}, \quad t \in \mathbb{R}
\]
The matrix \(\begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -7 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}\) has a solution set given by:
\[
\overrightarrow{\mathbf{x}} = s \begin{bmatrix} -2 \\ 7 \\ 1 \\ 0 \end{bmatrix}, \quad s \in \mathbb{R}
\]
Matrix (a):
- RREF: \(\boxed{\text{yes}}\)
- Determinant: \(\boxed{2}\)
- Solutions: \(\boxed{\overrightarrow{\mathbf{x}} = \overrightarrow{\mathbf{0}}}\)
Matrix (b):
- RREF: \(\boxed{\begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}}\)
- Determinant: \(\boxed{0}\)
- Solutions: \(\boxed{\overrightarrow{\mathbf{x}} = t \begin{bmatrix} -4 \\ 2 \\ 1 \end{bmatrix}, \, t \in \mathbb{R}}\)
Matrix (c):
- RREF: \(\boxed{\text{yes}}\)
- Determinant: \(\boxed{\text{not defined}}\)
- Solutions: \(\boxed{\overrightarrow{\mathbf{x}} = s \begin{bmatrix} -2 \\ 7 \\ 1 \\ 0 \end{bmatrix}, \, s \in \mathbb{R}}\)
![For each given matrix:
(i) Determine whether the matrix is in reduced row echelon form. If so, just write yes. If not, give the reduced row echelon form of the matrix.
(ii) Determine the determinant of the matrix.
(iii) Find all solutions to A x = 0 where A is the given matrix.
(a) [1 3 4; 0 2 1; 0 0 1]
(b) [1 0 4; 0 0 0; 0 1 -2]
(c) [1 0 2 0; 0 1 -7 0; 0 0 0 1; 0 0 0 0]](https://thumbnail.justsolvely.com/seoQuestionThumb/577a310c-bec4-4f6a-9525-54c32b58c5a1.webp)
Answered
