Math Problem Statement
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Let matrix A = \begin{bmatrix} 3 & -2 & 1 \\ 0 & 5 & 4 \\ -6 & 2 & -1 \end{bmatrix}. Define matrix B as the transpose of the transpose of the transpose of matrix A^T. Using row reduction, find the determinant of matrix A + B.
Solution
To solve this problem, let's carefully analyze and compute step-by-step:
Step 1: Define Matrix
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3 & -2 & 1 \\ 0 & 5 & 4 \\ -6 & 2 & -1 \end{bmatrix}.$$ ### Step 2: Define Matrix $$B$$ Matrix $$B$$ is defined as the **transpose of the transpose of the transpose** of matrix $$A^T$$. Since transposing a matrix three times is equivalent to just taking the transpose once, we conclude: $$B = A^T.$$ Thus, compute $$A^T$$ (the transpose of $$A$$): $$A^T = \begin{bmatrix} 3 & 0 & -6 \\ -2 & 5 & 2 \\ 1 & 4 & -1 \end{bmatrix}.$$ So, \( B = A^T = \begin{bmatrix} 3 & 0 & -6 \\ -2 & 5 & 2 \\ 1 & 4 & -1 \end{bmatrix}. \] ### Step 3: Compute $$A + B$$ Adding $$A$$ and $$B$$: $$A + B = \begin{bmatrix} 3 & -2 & 1 \\ 0 & 5 & 4 \\ -6 & 2 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 0 & -6 \\ -2 & 5 & 2 \\ 1 & 4 & -1 \end{bmatrix}.$$ Perform element-wise addition: $$A + B = \begin{bmatrix} 3+3 & -2+0 & 1+(-6) \\ 0+(-2) & 5+5 & 4+2 \\ -6+1 & 2+4 & -1+(-1) \end{bmatrix} = \begin{bmatrix} 6 & -2 & -5 \\ -2 & 10 & 6 \\ -5 & 6 & -2 \end{bmatrix}.$$ ### Step 4: Find the Determinant of $$A + B$$ using Row Reduction The matrix is: $$A + B = \begin{bmatrix} 6 & -2 & -5 \\ -2 & 10 & 6 \\ -5 & 6 & -2 \end{bmatrix}.$$ To find the determinant, perform row reduction while keeping track of operations. #### Row Reduction: 1. Use the first element (pivot = 6) in Row 1 to eliminate elements in column 1 of Row 2 and Row 3. - Row 2: $$R_2 \to R_2 + \frac{2}{6} R_1 = R_2 + \frac{1}{3} R_1$$, - Row 3: $$R_3 \to R_3 + \frac{5}{6} R_1$$. New matrix: $$\begin{bmatrix} 6 & -2 & -5 \\ 0 & \frac{32}{3} & \frac{28}{3} \\ 0 & \frac{4}{3} & -\frac{32}{3} \end{bmatrix}.$$ 2. Use the pivot in Row 2 ($$\frac{32}{3}$$) to eliminate the element in column 2 of Row 3. - Row 3: $$R_3 \to R_3 - \frac{\frac{4}{3}}{\frac{32}{3}} R_2 = R_3 - \frac{1}{8} R_2$$. New matrix: $$\begin{bmatrix} 6 & -2 & -5 \\ 0 & \frac{32}{3} & \frac{28}{3} \\ 0 & 0 & -\frac{40}{3} \end{bmatrix}.$$ 3. Determinant is the product of the diagonal elements: $$\det(A + B) = 6 \cdot \frac{32}{3} \cdot \left(-\frac{40}{3}\right).$$ #### Final Calculation: $$\det(A + B) = 6 \cdot \frac{32}{3} \cdot -\frac{40}{3} = -\frac{6 \cdot 32 \cdot 40}{9} = -\frac{7680}{9}.$$ ### Final Answer: $$\det(A + B) = -\frac{7680}{9}.$$ --- Would you like detailed steps on a specific part? Here are some related questions for further practice: 1. What is the determinant of $$A^T$$ directly? 2. How does row reduction affect the determinant during calculations? 3. Can you verify $$\det(A + B)$$ by cofactor expansion instead? 4. What are the eigenvalues of $$A + B$$? 5. Is $$A + B$$ invertible? Why or why not? **Tip:** Always double-check row reduction steps to avoid small arithmetic errors, especially with fractions!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Operations
Determinants
Row Reduction
Matrix Transpose
Formulas
Matrix Transpose: (A^T)^T = A
Determinant of a Matrix
Row Reduction: Elimination to upper triangular form
Theorems
Properties of Transpose Matrices
Determinant Invariance under Row Operations
Suitable Grade Level
Undergraduate Linear Algebra
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