Math Problem Statement
Find the determinant by row reduction to echelon form.
StartAbsoluteValue Start 4 By 4 Matrix 1st Row 1st Column 1 2nd Column negative 1 3rd Column negative 3 4st Column 0 2nd Row 1st Column 4 2nd Column negative 3 3rd Column 3 4st Column 2 3rd Row 1st Column 1 2nd Column 2 3rd Column 0 4st Column 3 4st Row 1st Column negative 2 2nd Column 5 3rd Column 9 4st Column 3 EndMatrix EndAbsoluteValue
1
−1
−3
0
4
−3
3
2
1
2
0
3
−2
5
9
3
Question content area bottom
Part 1
Use row operations to reduce the matrix to echelon form.
Start 4 By 4 Table 1st Row 1st Column 1 2nd Column negative 1 3rd Column negative 3 4st Column 0 2nd Row 1st Column 4 2nd Column negative 3 3rd Column 3 4st Column 2 3rd Row 1st Column 1 2nd Column 2 3rd Column 0 4st Column 3 4st Row 1st Column negative 2 2nd Column 5 3rd Column 9 4st Column 3 EndTable
1
−1
−3
0
4
−3
3
2
1
2
0
3
−2
5
9
3
~ enter your response here
Solution
We are tasked with finding the determinant of a matrix by reducing it to row echelon form. The given matrix is:
1 & -1 & -3 & 0 \\ 4 & -3 & 3 & 2 \\ 1 & 2 & 0 & 3 \\ -2 & 5 & 9 & 3 \end{pmatrix}$$ ### Step 1: Convert the matrix to echelon form using row operations. We will perform elementary row operations to make the entries below the main diagonal zero. #### Operation 1: Make the entries below $$a_{11}$$ zero - Subtract 4 times the first row from the second row: $$R_2 \rightarrow R_2 - 4R_1$$ $$R_2 = (4, -3, 3, 2) - 4 \times (1, -1, -3, 0) = (4 - 4, -3 + 4, 3 + 12, 2 - 0) = (0, 1, 15, 2)$$ - Subtract 1 times the first row from the third row: $$R_3 \rightarrow R_3 - R_1$$ $$R_3 = (1, 2, 0, 3) - (1, -1, -3, 0) = (1 - 1, 2 + 1, 0 + 3, 3 - 0) = (0, 3, 3, 3)$$ - Add 2 times the first row to the fourth row: $$R_4 \rightarrow R_4 + 2R_1$$ $$R_4 = (-2, 5, 9, 3) + 2 \times (1, -1, -3, 0) = (-2 + 2, 5 - 2, 9 - 6, 3 + 0) = (0, 3, 3, 3)$$ The matrix now becomes: $$\begin{pmatrix} 1 & -1 & -3 & 0 \\ 0 & 1 & 15 & 2 \\ 0 & 3 & 3 & 3 \\ 0 & 3 & 3 & 3 \end{pmatrix}$$ #### Operation 2: Make the entries below $$a_{22}$$ zero - Subtract 3 times the second row from the third row: $$R_3 \rightarrow R_3 - 3R_2$$ $$R_3 = (0, 3, 3, 3) - 3 \times (0, 1, 15, 2) = (0, 3 - 3, 3 - 45, 3 - 6) = (0, 0, -42, -3)$$ - Subtract 3 times the second row from the fourth row: $$R_4 \rightarrow R_4 - 3R_2$$ $$R_4 = (0, 3, 3, 3) - 3 \times (0, 1, 15, 2) = (0, 3 - 3, 3 - 45, 3 - 6) = (0, 0, -42, -3)$$ The matrix now becomes: $$\begin{pmatrix} 1 & -1 & -3 & 0 \\ 0 & 1 & 15 & 2 \\ 0 & 0 & -42 & -3 \\ 0 & 0 & -42 & -3 \end{pmatrix}$$ #### Operation 3: Make the entry below $$a_{33}$$ zero - Subtract the third row from the fourth row: $$R_4 \rightarrow R_4 - R_3$$ $$R_4 = (0, 0, -42, -3) - (0, 0, -42, -3) = (0, 0, 0, 0)$$ The matrix is now in echelon form: $$\begin{pmatrix} 1 & -1 & -3 & 0 \\ 0 & 1 & 15 & 2 \\ 0 & 0 & -42 & -3 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ ### Step 2: Calculate the determinant The determinant of an upper triangular matrix (in echelon form) is the product of the diagonal elements. $$\text{det}(A) = 1 \times 1 \times (-42) \times 0 = 0$$ Thus, the determinant of the matrix is $$0$$. --- Would you like more details or have any questions? Here are 5 related questions you could explore: 1. What does a zero determinant tell you about the matrix? 2. How can row reduction help in solving systems of linear equations? 3. What are the conditions under which a matrix has a non-zero determinant? 4. How does cofactor expansion differ from row reduction for finding determinants? 5. Can you describe the properties of an invertible matrix? **Tip:** If the determinant of a matrix is zero, the matrix is singular and not invertible, meaning it doesn't have a unique solution in systems of equations.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Linear Algebra
Determinants
Row Echelon Form
Matrix Operations
Formulas
Determinant of an upper triangular matrix is the product of the diagonal elements
Theorems
Elementary Row Operations
Determinant Theorem
Suitable Grade Level
Undergraduate (Linear Algebra Course) or Advanced High School (Grades 11-12)