Math Problem Statement
Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation. StartAbsoluteValue Start 4 By 4 Matrix 1st Row 1st Column 9 2nd Column 0 3rd Column 0 4st Column 5 2nd Row 1st Column 1 2nd Column 7 3rd Column 2 4st Column negative 3 3rd Row 1st Column 3 2nd Column 0 3rd Column 0 4st Column 0 4st Row 1st Column 2 2nd Column 3 3rd Column 1 4st Column 1 EndMatrix EndAbsoluteValue
Solution
To compute the determinant of the given matrix using cofactor expansion, we will choose a row or column that minimizes computation, which means selecting one with the most zeros.
The matrix is:
9 & 0 & 0 & 5 \\ 1 & 7 & 2 & -3 \\ 3 & 0 & 0 & 0 \\ 2 & 3 & 1 & 1 \\ \end{vmatrix}$$ We can expand along the third row since it contains the most zeros. ### Cofactor expansion along the third row: $$\text{det}(A) = 3 \cdot C_{31} + 0 \cdot C_{32} + 0 \cdot C_{33} + 0 \cdot C_{34}$$ Only the first term remains (because of the zeros in the third row): $$\text{det}(A) = 3 \cdot C_{31}$$ Where $$C_{31}$$ is the cofactor of the element at position (3,1). The cofactor is calculated as: $$C_{31} = (-1)^{3+1} \cdot \text{det of the } 3 \times 3 \text{ submatrix formed by deleting row 3 and column 1}$$ The submatrix formed by deleting row 3 and column 1 is: $$\begin{pmatrix} 0 & 0 & 5 \\ 7 & 2 & -3 \\ 3 & 1 & 1 \end{pmatrix}$$ Now, we calculate the determinant of this $$3 \times 3$$ submatrix by expanding along the first row (which has two zeros): $$\text{det} \begin{pmatrix} 0 & 0 & 5 \\ 7 & 2 & -3 \\ 3 & 1 & 1 \end{pmatrix} = 0 \cdot C_{11} + 0 \cdot C_{12} + 5 \cdot C_{13}$$ Only the last term remains, so we calculate $$C_{13}$$: $$C_{13} = (-1)^{1+3} \cdot \text{det of the } 2 \times 2 \text{ submatrix} \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$$ The determinant of the $$2 \times 2$$ submatrix is: $$\text{det} \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix} = (7 \cdot 1) - (2 \cdot 3) = 7 - 6 = 1$$ So, $$C_{13} = (-1)^{1+3} \cdot 1 = 1$$. Therefore, the determinant of the $$3 \times 3$$ submatrix is: $$\text{det} \begin{pmatrix} 0 & 0 & 5 \\ 7 & 2 & -3 \\ 3 & 1 & 1 \end{pmatrix} = 5 \cdot 1 = 5$$ ### Final determinant calculation: $$\text{det}(A) = 3 \cdot 5 = 15$$ Thus, the determinant of the matrix is $$\boxed{15}$$. --- Would you like more details on any step or have any questions? Here are 5 related questions you could explore: 1. What is cofactor expansion in a general sense for any matrix size? 2. How is a cofactor defined mathematically? 3. How does choosing a row or column with the most zeros simplify calculations? 4. What other methods exist for finding the determinant of a $$4 \times 4$$ matrix? 5. What are the applications of determinants in linear algebra? **Tip:** Always look for rows or columns with zeros when calculating determinants using cofactor expansion to minimize computation.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Linear Algebra
Determinants
Cofactor Expansion
Formulas
det(A) = Σ a_{ij} * C_{ij}
Cofactor C_{ij} = (-1)^{i+j} * det(submatrix)
Theorems
Cofactor Expansion Theorem
Determinant Properties
Suitable Grade Level
Grades 10-12