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Find the inverse of a using gauss elimination method

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

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

The augmented matrix is:

\[
\left[\begin{array}{ccc|ccc}
2 & 0 & 0 & 1 & 0 & 0 \\
2 & 1 & 0 & 0 & 1 & 0 \\
3 & 2 & 2 & 0 & 0 & 1
\end{array}\right]
\]

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}


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

Learn how to find the inverse of a 3x3 matrix A = [[2, 0, 0], [2, 1, 0], [3, 2, 2]] using Gauss-Jordan elimination. This step-by-step solution demonstrates key matrix algebra concepts like row operations and matrix inversion. Suitable for undergraduate students.

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