https://file.aiquickdraw.com/mathbot/file/1728322387577z5j31w9o.jpg

To solve the inverse of matrix \( A^{-1} \) for both matrices given in the image, we will use the **Gauss-Jordan elimination method**, which involves transforming the augmented matrix \( [A|I] \) into \( [I|A^{-1}] \).

---

### a) For matrix \( A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \):

1. **Step 1**: Write the augmented matrix \( [A | I] \), where \( I \) is the identity matrix:
   \[
   \begin{pmatrix}
   1 & 1 & 1 & | & 1 & 0 & 0 \\
   0 & 1 & 1 & | & 0 & 1 & 0 \\
   0 & 0 & 1 & | & 0 & 0 & 1 \\
   \end{pmatrix}
   \]

2. **Step 2**: Apply row operations to turn the left-hand side into the identity matrix.

   - Row 1 is already fine. Focus on Row 2.
   - Subtract Row 2 from Row 1:
     \[
     R1' = R1 - R2 \implies \begin{pmatrix} 1 & 0 & 0 & | & 1 & -1 & 1 \end{pmatrix}
     \]
   
   Thus, the final matrix becomes:
   \[
   \begin{pmatrix}
   1 & 0 & 0 & | & 1 & -1 & 1 \\
   0 & 1 & 0 & | & 0 & 1 & -1 \\
   0 & 0 & 1 & | & 0 & 0 & 1
   \end{pmatrix}
   \]

3. **Conclusion**: The inverse of matrix \( A \) is:
   \[
   A^{-1} = \begin{pmatrix}
   1 & -1 & 1 \\
   0 & 1 & -1 \\
   0 & 0 & 1
   \end{pmatrix}
   \]

---

### b) For matrix \( A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 3 & 4 & 0 \end{pmatrix} \):

1. **Step 1**: Write the augmented matrix \( [A | I] \):
   \[
   \begin{pmatrix}
   1 & 1 & 1 & | & 1 & 0 & 0 \\
   2 & 3 & 1 & | & 0 & 1 & 0 \\
   3 & 4 & 0 & | & 0 & 0 & 1
   \end{pmatrix}
   \]

2. **Step 2**: Apply row operations to transform \( A \) into the identity matrix.

   - First, subtract 2 times Row 1 from Row 2:
     \[
     R2' = R2 - 2R1 = \begin{pmatrix} 0 & 1 & -1 & | & -2 & 1 & 0 \end{pmatrix}
     \]
   
   - Subtract 3 times Row 1 from Row 3:
     \[
     R3' = R3 - 3R1 = \begin{pmatrix} 0 & 1 & -3 & | & -3 & 0 & 1 \end{pmatrix}
     \]
   
   - Now subtract Row 2 from Row 3 to get:
     \[
     R3'' = R3' - R2' = \begin{pmatrix} 0 & 0 & -2 & | & -1 & -1 & 1 \end{pmatrix}
     \]
   
   - Divide Row 3 by -2:
     \[
     R3''' = \begin{pmatrix} 0 & 0 & 1 & | & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \end{pmatrix}
     \]
   
   - Add 1 times Row 3 to Row 2:
     \[
     R2'' = \begin{pmatrix} 0 & 1 & 0 & | & -\frac{3}{2} & \frac{1}{2} & -\frac{1}{2} \end{pmatrix}
     \]
   
   - Subtract 1 times Row 3 from Row 1:
     \[
     R1'' = \begin{pmatrix} 1 & 0 & 0 & | & \frac{5}{2} & -\frac{1}{2} & \frac{1}{2} \end{pmatrix}
     \]

3. **Conclusion**: The inverse of matrix \( A \) is:
   \[
   A^{-1} = \begin{pmatrix}
   \frac{5}{2} & -\frac{1}{2} & \frac{1}{2} \\
   -\frac{3}{2} & \frac{1}{2} & -\frac{1}{2} \\
   \frac{1}{2} & \frac{1}{2} & -\frac{1}{2}
   \end{pmatrix}
   \]

---

Let me know if you'd like any details about the steps or if you have further questions! 

### Related questions:
1. How is the determinant of a matrix related to its invertibility?
2. Can a matrix with a determinant of 0 have an inverse? Why or why not?
3. What is the significance of row reduction in finding the inverse of a matrix?
4. How does the Gauss-Jordan elimination method differ from other methods of finding the inverse?
5. Can non-square matrices have inverses? Why or why not?

**Tip**: Always check if a matrix is invertible by calculating its determinant before attempting to find its inverse!

Tìm ma trận nghịch đảo A^{-1} bằng phép biến đổi sơ cấp theo hàng cho ma trận A đã cho

This page explains how to find the inverse of a 3x3 matrix using Gauss-Jordan elimination. The problem provides two matrices and demonstrates row reduction techniques to transform the augmented matrix [A|I] into [I|A^{-1}]. Suitable for undergraduate students studying linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation