Math Problem Statement
Question content area top Part 1 Find Upper A Superscript negative 1, or explain why it does not exist. Aequalsleft bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 1 2nd Row 1st Column 2 2nd Column 1 3rd Column negative 1 3rd Row 1st Column 5 2nd Column 1 3rd Column 0 EndMatrix right bracket
Question content area bottom Part 1 Select the correct choice below and fill in any answer boxes present in your choice. A. Upper A Superscript negative 1equals enter your response here (Simplify your answers.) B. Upper A Superscript negative 1 does not exist because A in the augmented matrix [A|I] cannot be transformed to I using the Gauss-Jordan method.
Solution
To find , we need to determine if matrix is invertible. If it is, we can calculate the inverse; if not, we will explain why the inverse does not exist.
Matrix is:
1 & 1 & 1 \\ 2 & 1 & -1 \\ 5 & 1 & 0 \end{bmatrix}$$ To check whether the matrix $$A$$ is invertible, we need to calculate its **determinant**. A matrix is invertible if and only if its determinant is non-zero. ### Step 1: Calculate the determinant of $$A$$ The determinant of a $$3 \times 3$$ matrix is given by: $$\text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)$$ For our matrix $$A$$, we assign: $$a = 1, b = 1, c = 1, d = 2, e = 1, f = -1, g = 5, h = 1, i = 0$$ Substitute these values into the formula: \[ \text{det}(A) = 1[(1 \cdot 0) - (-1 \cdot 1)] - 1[(2 \cdot 0) - (-1 \cdot 5)] + 1[(2 \cdot 1) - (1 \cdot 5)] \] \[ \text{det}(A) = 1[0 + 1] - 1[0 + 5] + 1[2 - 5] \] $$\text{det}(A) = 1 - 5 - 3$$ $$\text{det}(A) = -7$$ ### Step 2: Conclusion Since $$\text{det}(A) = -7$$, which is not equal to zero, matrix $$A$$ **is invertible**. ### Step 3: Inverse of $$A$$ Now, let's find $$A^{-1}$$. The inverse of a matrix $$A$$ is given by: $$A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)$$ Where $$\text{adj}(A)$$ is the adjugate (transpose of the cofactor matrix). Let's calculate that. The determinant of matrix $$A$$ is indeed $$-7$$, confirming that the matrix is invertible. The inverse of matrix $$A$$, $$A^{-1}$$, is: $$A^{-1} = \frac{1}{-7} \times \begin{bmatrix} 1 & 5 & -6 \\ 1 & 5 & -4 \\ -1 & 2 & -1 \end{bmatrix}$$ Thus, $$A^{-1}$$ is: $$A^{-1} = \begin{bmatrix} -1/7 & -5/7 & 6/7 \\ -1/7 & -5/7 & 4/7 \\ 1/7 & -2/7 & 1/7 \end{bmatrix}$$ Would you like further details or explanations? ### Related Questions: 1. How do you find the determinant of larger matrices? 2. What is the Gauss-Jordan method, and how does it relate to finding matrix inverses? 3. What happens when the determinant of a matrix is zero? 4. How can you verify if a matrix is invertible using row reduction? 5. What are some applications of matrix inverses in real-world problems? **Tip**: When calculating matrix inverses by hand, always check the determinant first to avoid unnecessary computations if it's zero!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Inversion
Determinants
Formulas
Determinant of a 3x3 matrix: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Inverse of a matrix: A^(-1) = (1/det(A)) * adj(A)
Theorems
Invertible Matrix Theorem
Suitable Grade Level
Grades 10-12, College