Math Problem Statement

Find the inverse of a using gauss elimination method

Find the inverse of the matrix A = [[2, 0, 0], [2, 1, 0], [3, 2, 2]] using Gauss-Jordan elimination.

Solution

To find the inverse of matrix A=[200210322]A = \begin{bmatrix} 2 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 2 \end{bmatrix} using Gauss-Jordan elimination, we proceed as follows:

We augment the matrix AA with the identity matrix and perform row operations to convert AA into the identity matrix. The resulting augmented part will be the inverse of AA.

The augmented matrix is:

2 & 0 & 0 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 3 & 2 & 2 & 0 & 0 & 1 \end{array}\right]$$ We perform row operations to transform the left side into the identity matrix. ### Step-by-Step Procedure: 1. **Row 1: Make the pivot in the first column a 1.** $$R_1 \rightarrow \frac{1}{2} R_1 = \left[1 \, 0 \, 0 \mid \frac{1}{2} \, 0 \, 0\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 3 & 2 & 2 & 0 & 0 & 1 \end{array}\right]$$ 2. **Row 2: Eliminate the first element in row 2.** Subtract 2 times row 1 from row 2: $$R_2 \rightarrow R_2 - 2R_1 = \left[0 \, 1 \, 0 \mid -1 \, 1 \, 0\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & 0 \\ 3 & 2 & 2 & 0 & 0 & 1 \end{array}\right]$$ 3. **Row 3: Eliminate the first element in row 3.** Subtract 3 times row 1 from row 3: $$R_3 \rightarrow R_3 - 3R_1 = \left[0 \, 2 \, 2 \mid -\frac{3}{2} \, 0 \, 1\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & 0 \\ 0 & 2 & 2 & -\frac{3}{2} & 0 & 1 \end{array}\right]$$ 4. **Row 3: Make the pivot in the second column a 1.** Divide row 3 by 2: $$R_3 \rightarrow \frac{1}{2}R_3 = \left[0 \, 1 \, 1 \mid -\frac{3}{4} \, 0 \, \frac{1}{2}\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & 0 \\ 0 & 1 & 1 & -\frac{3}{4} & 0 & \frac{1}{2} \end{array}\right]$$ 5. **Row 2: Eliminate the second element in row 2.** Subtract row 3 from row 2: $$R_2 \rightarrow R_2 - R_3 = \left[0 \, 0 \, -1 \mid -\frac{1}{4} \, 1 \, -\frac{1}{2}\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & -1 & -\frac{1}{4} & 1 & -\frac{1}{2} \\ 0 & 1 & 1 & -\frac{3}{4} & 0 & \frac{1}{2} \end{array}\right]$$ 6. **Row 2: Make the pivot in the third column a 1.** Multiply row 2 by -1: $$R_2 \rightarrow -1R_2 = \left[0 \, 0 \, 1 \mid \frac{1}{4} \, -1 \, \frac{1}{2}\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{4} & -1 & \frac{1}{2} \\ 0 & 1 & 1 & -\frac{3}{4} & 0 & \frac{1}{2} \end{array}\right]$$ 7. **Row 3: Eliminate the third element in row 3.** Subtract row 2 from row 3: $$R_3 \rightarrow R_3 - R_2 = \left[0 \, 1 \, 0 \mid -1 \, 1 \, 0\right]$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{4} & -1 & \frac{1}{2} \\ 0 & 1 & 0 & -1 & 1 & 0 \end{array}\right]$$ 8. **Row 2: Eliminate the third element in row 2.** Add $$\frac{1}{4}$$ times row 3 to row 2: $$R_2 \rightarrow R_2 + \frac{1}{4} R_3 = \left[1 \, 0 \, 0 \mid \frac{3}{2} \, -2 \, 1\right]$$ After all these operations, the inverse matrix of $$A$$ is: \[ A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ -1 & 1 & 0 \\ \frac{3}{4} & 0 & 1 \end{bmatrix}

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
Gauss-Jordan Elimination

Formulas

Gauss-Jordan Elimination for Matrix Inversion
Elementary Row Operations

Theorems

Inverse of a Matrix

Suitable Grade Level

Undergraduate

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