Use the elementary matrix \( E \) to find \( E A \) using an elementary row operation. \( E=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right] \quad A=\left[\begin{array}{ccc}-1 & -2 & -5 \\ -6 & -8 & 2 \\ 5 & -9 & 6\end{array}\right] \) \( E A=\left[\begin{array}{l}\text { Ex:5 } \\ \square\end{array}\right] \)

To find \( E A \), we perform the matrix multiplication of the elementary matrix \( E \) and the matrix \( A \). The matrix \( E \) swaps the second and third rows of any matrix it multiplies. Calculating \( E A \): \[ E A = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right] \left[\begin{array}{ccc}-1 & -2 & -5 \\ -6 & -8 & 2 \\ 5 & -9 & 6\end{array}\right] \] By performing the multiplication, we find: 1. The first row of \( E A \) is the first row of \( A \): \([-1, -2, -5]\). 2. The second row of \( E A \) takes the third row of \( A \): \([5, -9, 6]\). 3. The third row of \( E A \) takes the second row of \( A \): \([-6, -8, 2]\). Putting it all together, we have: \[ E A = \left[\begin{array}{ccc}-1 & -2 & -5 \\ 5 & -9 & 6 \\ -6 & -8 & 2\end{array}\right] \] So, \( E A = \left[\begin{array}{ccc}-1 & -2 & -5 \\ 5 & -9 & 6 \\ -6 & -8 & 2\end{array}\right] \).