To find the inverse of the matrix $A$ using row operations, we'll perform Gaussian elimination on the augmented matrix $[A | I]$, where $I$ is the identity matrix. The goal is to transform the matrix $A$ into the identity matrix $I$. If successful, the right side of the augmented matrix will be the inverse of $A$.

Given matrix $A$:

1 & -2 & 0 & 2 \\ -1 & 3 & 1 & -2 \\ -1 & 5 & 3 & -2 \\ 0 & 7 & 7 & 0 \end{pmatrix}$$ We start by forming the augmented matrix $$[A | I]$$: $$[A | I] = \begin{pmatrix} 1 & -2 & 0 & 2 & | & 1 & 0 & 0 & 0 \\ -1 & 3 & 1 & -2 & | & 0 & 1 & 0 & 0 \\ -1 & 5 & 3 & -2 & | & 0 & 0 & 1 & 0 \\ 0 & 7 & 7 & 0 & | & 0 & 0 & 0 & 1 \end{pmatrix}$$ ### Step 1: Make the first pivot element 1 The first element of the first row is already 1, so we don't need to change the first row. ### Step 2: Eliminate below the first pivot Add $$R1$$ to $$R2$$ to eliminate the first element in the second row: $$R2 \rightarrow R2 + R1 = \begin{pmatrix} 0 & 1 & 1 & 0 & | & 0 & 1 & 0 & 0 \end{pmatrix}$$ Add $$R1$$ to $$R3$$ to eliminate the first element in the third row: $$R3 \rightarrow R3 + R1 = \begin{pmatrix} 0 & 3 & 3 & 0 & | & 0 & 0 & 1 & 0 \end{pmatrix}$$ No change is needed for $$R4$$ because its first element is already 0. The matrix now looks like this: $$\begin{pmatrix} 1 & -2 & 0 & 2 & | & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & | & 1 & 1 & 0 & 0 \\ 0 & 3 & 3 & 0 & | & 1 & 0 & 1 & 0 \\ 0 & 7 & 7 & 0 & | & 0 & 0 & 0 & 1 \end{pmatrix}$$ ### Step 3: Make the second pivot element 1 The second pivot element is already 1 in $$R2$$. ### Step 4: Eliminate below the second pivot Subtract $$3 \times R2$$ from $$R3$$: $$R3 \rightarrow R3 - 3 \times R2 = \begin{pmatrix} 0 & 0 & 0 & 0 & | & -3 & -3 & 1 & 0 \end{pmatrix}$$ Subtract $$7 \times R2$$ from $$R4$$: $$R4 \rightarrow R4 - 7 \times R2 = \begin{pmatrix} 0 & 0 & 0 & 0 & | & -7 & -7 & 0 & 1 \end{pmatrix}$$ Now the matrix is: $$\begin{pmatrix} 1 & -2 & 0 & 2 & | & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & | & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & | & -3 & -3 & 1 & 0 \\ 0 & 0 & 0 & 0 & | & -7 & -7 & 0 & 1 \end{pmatrix}$$ ### Step 5: Check for further pivot elements At this stage, we see that the third and fourth rows have zeros in all elements of $$A$$, but not in the corresponding identity columns, indicating that the matrix is **singular**. Since we have a row of zeros in the coefficient matrix $$A$$ without a corresponding row of zeros in the identity matrix, $$A$$ does **not** have an inverse. ### Conclusion The matrix $$A$$ does not have an inverse. --- Do you want more details on any of the steps, or have any questions? Here are some follow-up questions: 1. How can you determine if a matrix is invertible without row operations? 2. What does it mean when a matrix is singular? 3. How can you apply row operations to solve a system of equations? 4. What are some other methods for finding the inverse of a matrix? 5. How can you check your answer after finding the inverse of a matrix? **Tip:** A square matrix has an inverse only if its determinant is non-zero.