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 use the elementary matrix E to perform an elementary row operation on A, we multiply E by A. Here,  E = [ [1, 0, 0],     [0, 0, 1],     [0, 1, 0] ] and  A = [ [ -1, -2, -5 ],     [ -6, -8, 2 ],     [ 5, -9, 6 ] ]. Note that E is the elementary row operation matrix that swaps the 2nd and 3rd rows. In other words, E acting on A swaps A’s second and third rows. Step by step, here’s how EA is computed: 1. Multiply E by A:   EA =    [ (Row 1 of E) · A,     (Row 2 of E) · A,     (Row 3 of E) · A ]. 2. Since the first row of E is [1, 0, 0], the first row of EA is equal to the first row of A:   Row 1 of EA = [ -1, -2, -5 ]. 3. The second row of E is [0, 0, 1]. Multiplying this row with A selects A's third row:   Row 2 of EA = [ 5, -9, 6 ]. 4. The third row of E is [0, 1, 0]. Multiplying this row with A selects A's second row:   Row 3 of EA = [ -6, -8, 2 ]. Thus, the product EA is:  EA = [ [ -1, -2, -5 ],      [ 5, -9, 6 ],      [ -6, -8, 2 ] ].