Math Problem Statement
what is the inverse of a matrix with column 1(1,-1,1) column 2(1,0,0) and column 3(1,1,2) using the gaussian elimination method
Solution
To find the inverse of a matrix using Gaussian elimination, follow these steps:
1. Write the matrix and an identity matrix side by side.
Let the matrix be:
1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}.$$ Create an augmented matrix by appending the identity matrix $$I$$ of the same size: $$\left[ A | I \right] = \begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 2 & 0 & 0 & 1 \end{bmatrix}.$$ ### 2. Apply row operations to convert the left part of the augmented matrix to the identity matrix. The goal is to transform $$\left[ A | I \right]$$ into $$\left[ I | A^{-1} \right]$$. #### Step 1: Make the first element of the first row 1 (already 1). The matrix remains: $$\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 2 & 0 & 0 & 1 \end{bmatrix}.$$ #### Step 2: Eliminate the first entry in the second and third rows. Add $$R_1$$ to $$R_2$$ and subtract $$R_1$$ from $$R_3$$: $$R_2 = R_2 + R_1, \quad R_3 = R_3 - R_1.$$ The new matrix is: $$\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & 1 & 1 & 0 \\ 0 & -1 & 1 & -1 & 0 & 1 \end{bmatrix}.$$ #### Step 3: Make the pivot in the second row 1 (already 1). Add $$R_2$$ to $$R_3$$ to eliminate the $$-1$$ in $$R_3$$, second column: $$R_3 = R_3 + R_2.$$ The new matrix is: $$\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & 1 & 1 & 0 \\ 0 & 0 & 3 & 0 & 1 & 1 \end{bmatrix}.$$ #### Step 4: Scale the third row to make the pivot 1. Divide $$R_3$$ by 3: $$R_3 = \frac{R_3}{3}.$$ The new matrix is: $$\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{3} & \frac{1}{3} \end{bmatrix}.$$ #### Step 5: Eliminate the third column from $$R_1$$ and $$R_2$$. For $$R_1$$: Subtract $$R_3$$ from $$R_1$$. For $$R_2$$: Subtract $$2 \times R_3$$ from $$R_2$$. $$R_1 = R_1 - R_3, \quad R_2 = R_2 - 2R_3.$$ The new matrix is: $$\begin{bmatrix} 1 & 1 & 0 & 1 & -\frac{1}{3} & -\frac{1}{3} \\ 0 & 1 & 0 & 1 & \frac{1}{3} & -\frac{2}{3} \\ 0 & 0 & 1 & 0 & \frac{1}{3} & \frac{1}{3} \end{bmatrix}.$$ #### Step 6: Eliminate the second column from $$R_1$$. Subtract $$R_2$$ from $$R_1$$: $$R_1 = R_1 - R_2.$$ The final matrix is: $$\begin{bmatrix} 1 & 0 & 0 & 0 & -\frac{2}{3} & \frac{1}{3} \\ 0 & 1 & 0 & 1 & \frac{1}{3} & -\frac{2}{3} \\ 0 & 0 & 1 & 0 & \frac{1}{3} & \frac{1}{3} \end{bmatrix}.$$ ### 3. Extract the inverse matrix. The right side of the augmented matrix is $$A^{-1}$$: $$A^{-1} = \begin{bmatrix} -\frac{2}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}.$$ ### Final Answer: $$A^{-1} = \begin{bmatrix} -\frac{2}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}.$$ Would you like a breakdown of any step in more detail or a verification of the result? --- **5 Related Questions:** 1. How is Gaussian elimination different from other methods for finding the inverse of a matrix? 2. Can every square matrix have an inverse? Why or why not? 3. How would you solve a system of equations using an inverse matrix? 4. What is the computational complexity of finding an inverse using Gaussian elimination? 5. How does row reduction help in solving linear equations? **Tip:** Always check whether a matrix is invertible by calculating its determinant. If $$\det(A) = 0$$, the matrix has no inverse!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Matrix Inversion
Gaussian Elimination
Formulas
Inverse of a matrix: A^-1 such that A * A^-1 = I
Gaussian elimination for solving augmented matrix
Theorems
Invertibility Theorem
Properties of Identity Matrices
Suitable Grade Level
Grades 11-12, Undergraduate
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