Math Problem Statement

what is inverse of this 3x3 matrix (im listing first a11 then a12...) :1,1,2,1,0,1,-1,1,-4

Solution

To find the inverse of the given 3×33 \times 3 matrix:

1 & 1 & 2 \\ 1 & 0 & 1 \\ -1 & 1 & -4 \end{pmatrix}$$ We can use the formula for the inverse of a matrix, which is given by: $$A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)$$ ### Step 1: Calculate the determinant of $$A$$ The determinant of a $$3 \times 3$$ matrix is calculated as follows: $$\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ For our matrix $$A$$: - $$a = 1, b = 1, c = 2$$ - $$d = 1, e = 0, f = 1$$ - $$g = -1, h = 1, i = -4$$ Calculating the determinant: $$\text{det}(A) = 1(0 \cdot -4 - 1 \cdot 1) - 1(1 \cdot -4 - 1 \cdot -1) + 2(1 \cdot 1 - 0 \cdot -1)$$ $$= 1(0 - 1) - 1(-4 + 1) + 2(1)$$ $$= -1 + 3 + 2 = 4$$ ### Step 2: Calculate the adjugate of $$A$$ The adjugate is the transpose of the cofactor matrix. The cofactor matrix is obtained by calculating the determinant of the $$2 \times 2$$ matrices formed by eliminating one row and one column. 1. **Cofactor of $$a_{11}$$**: $$\begin{vmatrix} 0 & 1 \\ 1 & -4 \end{vmatrix} = (0)(-4) - (1)(1) = -1$$ 2. **Cofactor of $$a_{12}$$**: $$\begin{vmatrix} 1 & 1 \\ -1 & -4 \end{vmatrix} = (1)(-4) - (1)(-1) = -4 + 1 = -3$$ 3. **Cofactor of $$a_{13}$$**: $$\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = (1)(1) - (0)(-1) = 1$$ 4. **Cofactor of $$a_{21}$$**: $$\begin{vmatrix} 1 & 2 \\ -1 & -4 \end{vmatrix} = (1)(-4) - (2)(-1) = -4 + 2 = -2$$ 5. **Cofactor of $$a_{22}$$**: $$\begin{vmatrix} 1 & 2 \\ -1 & -4 \end{vmatrix} = (1)(-4) - (2)(-1) = -4 + 2 = -2$$ 6. **Cofactor of $$a_{23}$$**: $$\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2$$ 7. **Cofactor of $$a_{31}$$**: $$\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1$$ 8. **Cofactor of $$a_{32}$$**: $$\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1$$ 9. **Cofactor of $$a_{33}$$**: $$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = (1)(0) - (1)(1) = -1$$ The cofactor matrix $$C$$ is: $$C = \begin{pmatrix} -1 & 3 & 1 \\ -2 & -2 & 2 \\ -1 & -1 & -1 \end{pmatrix}$$ ### Step 3: Transpose the cofactor matrix to get the adjugate $$\text{adj}(A) = C^T = \begin{pmatrix} -1 & -2 & -1 \\ 3 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix}$$ ### Step 4: Calculate the inverse Now we can find the inverse: $$A^{-1} = \frac{1}{4} \begin{pmatrix} -1 & -2 & -1 \\ 3 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{4} & -\frac{1}{2} & -\frac{1}{4} \\ \frac{3}{4} & -\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \end{pmatrix}$$ ### Final Result Thus, the inverse of the matrix is: $$A^{-1} = \begin{pmatrix} -\frac{1}{4} & -\frac{1}{2} & -\frac{1}{4} \\ \frac{3}{4} & -\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \end{pmatrix}$$ If you have any questions or need further details, let me know! Here are some related questions you might find interesting: 1. What is the geometric interpretation of a matrix's inverse? 2. How does the determinant relate to the invertibility of a matrix? 3. Can you explain the process of calculating the adjugate matrix in more detail? 4. What are the applications of inverse matrices in real-world scenarios? 5. How do you check if your calculated inverse is correct? **Tip:** When dealing with larger matrices, consider using numerical software for efficiency.

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
Determinants
Adjugate Matrices

Formulas

A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)

Theorems

Matrix Inversion Theorem

Suitable Grade Level

Grades 10-12