Compute the inverse using row reduction. \( \left[\begin{array}{cc}7 & -1 \\ 21 & 0\end{array}\right] \)

To find the inverse of the matrix \( A = \left[\begin{array}{cc}7 & -1 \\ 21 & 0\end{array}\right] \) using row reduction, we will augment the matrix \( A \) with the identity matrix \( I \): \[ \left[\begin{array}{cc|cc} 7 & -1 & 1 & 0 \\ 21 & 0 & 0 & 1 \end{array}\right] \] Next, we will perform row operations to transform the left side into the identity matrix. 1. **First, we can simplify the first row by dividing it by 7:** \[ R_1 \leftarrow \frac{1}{7} R_1 \implies \left[\begin{array}{cc|cc} 1 & -\frac{1}{7} & \frac{1}{7} & 0 \\ 21 & 0 & 0 & 1 \end{array}\right] \] 2. **Next, we will eliminate the first element of the second row by replacing \( R_2 \) with \( R_2 - 21R_1 \):** \[ R_2 \leftarrow R_2 - 21R_1 \implies \left[\begin{array}{cc|cc} 1 & -\frac{1}{7} & \frac{1}{7} & 0 \\ 0 & 3 & -3 & 1 \end{array}\right] \] 3. **Now, we simplify the second row by dividing it by 3:** \[ R_2 \leftarrow \frac{1}{3} R_2 \implies \left[\begin{array}{cc|cc} 1 & -\frac{1}{7} & \frac{1}{7} & 0 \\ 0 & 1 & -1 & \frac{1}{3} \end{array}\right] \] 4. **Next, we eliminate the second element of the first row by replacing \( R_1 \) with \( R_1 + \frac{1}{7} R_2 \):** \[ R_1 \leftarrow R_1 + \frac{1}{7} R_2 \implies \left[\begin{array}{cc|cc} 1 & 0 & \frac{1}{7} - \frac{1}{7} & \frac{1}{21} \\ 0 & 1 & -1 & \frac{1}{3} \end{array}\right] \] This simplifies to: \[ \left[\begin{array}{cc|cc} 1 & 0 & \frac{1}{7} & \frac{1}{21} \\ 0 & 1 & -1 & \frac{1}{3} \end{array}\right] \] 5. **Finally, we can express the inverse matrix from the augmented part:** The inverse of \( A \) is given by the right side of the augmented matrix: \[ A^{-1} = \left[\begin{array}{cc} \frac{1}{7} & \frac{1}{21} \\ -1 & \frac{1}{3} \end{array}\right] \] Thus, the inverse of the matrix \( A \) is: \[ A^{-1} = \left[\begin{array}{cc} \frac{1}{7} & \frac{1}{21} \\ -1 & \frac{1}{3} \end{array}\right] \]