To find the inverse of matrix $A = \begin{bmatrix} 2 & 1 & 1 \\ 3 & 2 & 3 \\ 1 & 4 & 9 \end{bmatrix}$ using the Gauss-Jordan Method, follow these steps:

Step 1: Form the Augmented Matrix

You start by writing the augmented matrix $[A | I]$, where $A$ is the given matrix, and $I$ is the identity matrix of the same size:

2 & 1 & 1 & 1 & 0 & 0 \\ 3 & 2 & 3 & 0 & 1 & 0 \\ 1 & 4 & 9 & 0 & 0 & 1 \\ \end{array}\right]$$ ### Step 2: Perform Row Operations to Transform $$A$$ into the Identity Matrix The goal is to use elementary row operations to transform the left side (matrix $$A$$) into the identity matrix, while applying the same operations to the right side. The right side will become the inverse of $$A$$. #### (1) Make the first pivot element in row 1 a 1: Divide row 1 by 2: $$R_1 \rightarrow \frac{1}{2} R_1$$ $$\left[\begin{array}{ccc|ccc} 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 3 & 2 & 3 & 0 & 1 & 0 \\ 1 & 4 & 9 & 0 & 0 & 1 \\ \end{array}\right]$$ #### (2) Eliminate the first column below the pivot: - $$R_2 \rightarrow R_2 - 3 \times R_1$$ - $$R_3 \rightarrow R_3 - R_1$$ $$\left[\begin{array}{ccc|ccc} 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & \frac{3}{2} & -\frac{3}{2} & 1 & 0 \\ 0 & \frac{7}{2} & \frac{17}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array}\right]$$ #### (3) Make the second pivot element a 1: Multiply row 2 by 2: $$R_2 \rightarrow 2 \times R_2$$ $$\left[\begin{array}{ccc|ccc} 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 3 & -3 & 2 & 0 \\ 0 & \frac{7}{2} & \frac{17}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array}\right]$$ #### (4) Eliminate the second column above and below the pivot: - $$R_1 \rightarrow R_1 - \frac{1}{2} \times R_2$$ - $$R_3 \rightarrow R_3 - \frac{7}{2} \times R_2$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & -1 & 2 & -1 & 0 \\ 0 & 1 & 3 & -3 & 2 & 0 \\ 0 & 0 & -2 & 10 & -7 & 1 \\ \end{array}\right]$$ #### (5) Make the third pivot element a 1: Multiply row 3 by $$-\frac{1}{2}$$: $$R_3 \rightarrow -\frac{1}{2} \times R_3$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & -1 & 2 & -1 & 0 \\ 0 & 1 & 3 & -3 & 2 & 0 \\ 0 & 0 & 1 & -5 & \frac{7}{2} & -\frac{1}{2} \\ \end{array}\right]$$ #### (6) Eliminate the third column above the pivot: - $$R_1 \rightarrow R_1 + R_3$$ - $$R_2 \rightarrow R_2 - 3 \times R_3$$ $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -3 & \frac{5}{2} & -\frac{1}{2} \\ 0 & 1 & 0 & 12 & -\frac{11}{2} & \frac{3}{2} \\ 0 & 0 & 1 & -5 & \frac{7}{2} & -\frac{1}{2} \\ \end{array}\right]$$ ### Step 3: The Inverse Matrix The right-hand side of the augmented matrix is now the inverse of $$A$$: $$A^{-1} = \begin{bmatrix} -3 & \frac{5}{2} & -\frac{1}{2} \\ 12 & -\frac{11}{2} & \frac{3}{2} \\ -5 & \frac{7}{2} & -\frac{1}{2} \end{bmatrix}$$ This is the inverse of the matrix $$A$$. --- Would you like further details on any of these steps, or do you have any questions? --- Here are five related questions to expand on this concept: 1. How does the Gauss-Jordan method compare to other methods for finding an inverse? 2. What happens if a matrix is not invertible in the Gauss-Jordan method? 3. Can you explain how row operations affect the solution when using augmented matrices? 4. How can you check if the inverse matrix calculated is correct? 5. How does the determinant of a matrix relate to its invertibility? **Tip:** A matrix is invertible if and only if its determinant is non-zero.