Math Problem Statement
inverse matrices of 2 1 3 7 4 2 5 3 10
Solution
To compute the inverse of a matrix, the following steps are used. Given the matrix:
2 & 1 & 3 \\ 7 & 4 & 2 \\ 5 & 3 & 10 \end{bmatrix}$$ ### Step 1: Check if the matrix is invertible. A matrix is invertible if its determinant is non-zero. The determinant of $$A$$ is calculated as follows: $$\text{det}(A) = 2(4 \cdot 10 - 2 \cdot 3) - 1(7 \cdot 10 - 2 \cdot 5) + 3(7 \cdot 3 - 4 \cdot 5)$$ Compute this: $$\text{det}(A) = 2(40 - 6) - 1(70 - 10) + 3(21 - 20)$$ $$\text{det}(A) = 2(34) - 1(60) + 3(1)$$ $$\text{det}(A) = 68 - 60 + 3 = 11$$ Since $$\text{det}(A) = 11 \neq 0$$, the matrix is invertible. ### Step 2: Compute the adjugate matrix ($$\text{adj}(A)$$). The adjugate is the transpose of the cofactor matrix. Each cofactor is the determinant of the $$2 \times 2$$ submatrix obtained by removing the row and column of the element. #### Minor determinants for the cofactors: - $$C_{11} = \text{det}\begin{bmatrix} 4 & 2 \\ 3 & 10 \end{bmatrix} = 4 \cdot 10 - 2 \cdot 3 = 40 - 6 = 34$$ - $$C_{12} = -\text{det}\begin{bmatrix} 7 & 2 \\ 5 & 10 \end{bmatrix} = -(7 \cdot 10 - 2 \cdot 5) = -(70 - 10) = -60$$ - $$C_{13} = \text{det}\begin{bmatrix} 7 & 4 \\ 5 & 3 \end{bmatrix} = 7 \cdot 3 - 4 \cdot 5 = 21 - 20 = 1$$ Continue computing cofactors for each element of $$A$$. The cofactor matrix is: $$\text{Cofactor}(A) = \begin{bmatrix} 34 & -60 & 1 \\ -16 & 20 & -1 \\ -9 & 29 & 2 \end{bmatrix}$$ #### Adjugate matrix ($$\text{adj}(A)$$): Transpose the cofactor matrix: $$\text{adj}(A) = \begin{bmatrix} 34 & -16 & -9 \\ -60 & 20 & 29 \\ 1 & -1 & 2 \end{bmatrix}$$ ### Step 3: Compute the inverse. The inverse is given by: $$A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)$$ Substitute $$\text{det}(A) = 11$$: $$A^{-1} = \frac{1}{11} \begin{bmatrix} 34 & -16 & -9 \\ -60 & 20 & 29 \\ 1 & -1 & 2 \end{bmatrix}$$ $$A^{-1} = \begin{bmatrix} \frac{34}{11} & \frac{-16}{11} & \frac{-9}{11} \\ \frac{-60}{11} & \frac{20}{11} & \frac{29}{11} \\ \frac{1}{11} & \frac{-1}{11} & \frac{2}{11} \end{bmatrix}$$ ### Final Answer: $$A^{-1} = \begin{bmatrix} \frac{34}{11} & \frac{-16}{11} & \frac{-9}{11} \\ \frac{-60}{11} & \frac{20}{11} & \frac{29}{11} \\ \frac{1}{11} & \frac{-1}{11} & \frac{2}{11} \end{bmatrix}$$ --- Let me know if you'd like details about any specific step or further assistance. #### 5 Questions to Explore: 1. What does the determinant of a matrix signify? 2. How can the adjugate matrix be used in other mathematical applications? 3. What are some properties of invertible matrices? 4. Can every square matrix be inverted? Why or why not? 5. How does the computational complexity of finding the inverse compare to other matrix operations? **Tip:** Always verify the determinant of a matrix before attempting to calculate its inverse—it saves time and avoids unnecessary calculations for singular matrices.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Determinants
Adjugate Matrices
Matrix Inverses
Formulas
Determinant of a 3x3 matrix: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Matrix inverse: A^(-1) = (1/det(A)) * adj(A)
Adjugate matrix: adj(A) = transpose of cofactor matrix
Theorems
Invertibility Criterion (det(A) ≠ 0)
Properties of Determinants
Suitable Grade Level
Grades 11-12, Undergraduate
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