Math Problem Statement
An alternative method for computing the inverse of a matrix π is described as follows: 1. Augment the matrix π with an identity matrix ππ of the same order. In the augmented matrix, place π on the left and ππ on the right. (i.e., π:ππ). 2. Using elementary row operations, try to convert the matrix on the left (where π was) into an identity matrix ππ. If successful, the matrix on the right should be the inverse matrix πβ1 (i.e., ππ:πβ1). 3. If an identity matrix cannot be formed on the left (e.g., a row of zeros appears), then π is singular and does not have an inverse. Using this method described above, determine the inverse of the following matrix, if it exists. [ 1 0 2 2 1 1 2 0 3
Solution
We will use the method you outlined to determine the inverse of the given matrix. The matrix is:
1 & 0 & 2 \\ 2 & 1 & 1 \\ 2 & 0 & 3 \end{bmatrix}$$ ### Step 1: Augment the matrix $$A$$ with the identity matrix $$I_3$$. The augmented matrix is: $$[A | I_3] = \begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 2 & 1 & 1 & | & 0 & 1 & 0 \\ 2 & 0 & 3 & | & 0 & 0 & 1 \end{bmatrix}$$ ### Step 2: Use elementary row operations to transform the left side of the augmented matrix into the identity matrix. #### Perform row operations: 1. Subtract $$2 \times$$ row 1 from row 2 to eliminate the first element of row 2. $$R_2 \rightarrow R_2 - 2R_1$$ $$\begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 2-2(1) & 1-2(0) & 1-2(2) & | & 0-2(1) & 1-2(0) & 0-2(0) \\ 2 & 0 & 3 & | & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 2 & 0 & 3 & | & 0 & 0 & 1 \end{bmatrix}$$ 2. Subtract $$2 \times$$ row 1 from row 3 to eliminate the first element of row 3. $$R_3 \rightarrow R_3 - 2R_1$$ $$\begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 2-2(1) & 0-2(0) & 3-2(2) & | & 0-2(1) & 0-2(0) & 1-2(0) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 0 & 0 & -1 & | & -2 & 0 & 1 \end{bmatrix}$$ 3. Multiply row 3 by $$-1$$ to make the pivot equal to 1. $$R_3 \rightarrow -R_3$$ $$\begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & -3 & | & -2 & 1 & 0 \\ 0 & 0 & 1 & | & 2 & 0 & -1 \end{bmatrix}$$ 4. Add $$3 \times$$ row 3 to row 2 to eliminate the third element of row 2. $$R_2 \rightarrow R_2 + 3R_3$$ $$\begin{bmatrix} 1 & 0 & 2 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & 4 & 1 & -3 \\ 0 & 0 & 1 & | & 2 & 0 & -1 \end{bmatrix}$$ 5. Subtract $$2 \times$$ row 3 from row 1 to eliminate the third element of row 1. $$R_1 \rightarrow R_1 - 2R_3$$ $$\begin{bmatrix} 1 & 0 & 0 & | & -3 & 0 & 2 \\ 0 & 1 & 0 & | & 4 & 1 & -3 \\ 0 & 0 & 1 & | & 2 & 0 & -1 \end{bmatrix}$$ Now, the left side of the augmented matrix is the identity matrix, and the right side is the inverse of $$A$$. ### Step 3: The inverse of $$A$$ is: $$A^{-1} = \begin{bmatrix} -3 & 0 & 2 \\ 4 & 1 & -3 \\ 2 & 0 & -1 \end{bmatrix}$$ ### Final Answer: The inverse of the matrix $$A$$ is: $$A^{-1} = \begin{bmatrix} -3 & 0 & 2 \\ 4 & 1 & -3 \\ 2 & 0 & -1 \end{bmatrix}$$ Would you like to explore further details or have any questions on this method? --- Here are five related questions you might find interesting: 1. How can we verify that the calculated inverse is correct? 2. What would happen if a row of zeros appeared during the row reduction process? 3. How does this method compare to finding the inverse using the formula for a 3x3 matrix? 4. What happens if the matrix is not invertible, and how can we identify this? 5. Can this method be applied to non-square matrices? **Tip:** Always check for row swaps and zero pivots during row reductionβthese can indicate issues with invertibility!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Inversion
Elementary Row Operations
Linear Algebra
Formulas
Inverse of a Matrix using Augmentation and Row Operations
Theorems
Invertibility of a Matrix
Row Operations and Inverses
Suitable Grade Level
Grades 11-12
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