The image contains a proof that a square matrix $A$ is invertible if and only if $\det(A) \neq 0$, involving the relationship between the determinant of the matrix and its reduced row echelon form $R$. The elementary steps in the proof go as follows:

1. The goal is to show that $\det(A) \neq 0$ if and only if $\det(R) \neq 0$.
2. $E_1, E_2, \dots, E_r$ are the elementary matrices corresponding to the row operations that reduce $A$ to $R$.
3. Therefore, $E_r E_{r-1} \dots E_1 A = R$, meaning these matrices transform $A$ into its reduced form $R$.
4. Taking the determinant of both sides: $\det(E_r) \det(E_{r-1}) \dots \det(E_1) \det(A) = \det(R)$
5. Since the determinants of the elementary matrices $E_i$ are non-zero (depending on the type of row operation), we can conclude that $\det(A) \neq 0$ if and only if $\det(R) \neq 0$.
6. Assuming $A$ is invertible, this implies that $\det(A) \neq 0$, and hence $R = I_n$, where $I_n$ is the identity matrix.
7. If $\det(A) = 0$, then $\det(R) = 0$, meaning $R$ cannot be invertible (it would have a row of zeros).
8. Therefore, $A$ is invertible if and only if $R = I_n$ and $\det(A) \neq 0$.

The fact that $\det(EA) = \det(E)\det(A)$ for an elementary matrix $E$ is used in step 5, where the determinants of both sides of the equation $E_r E_{r-1} \dots E_1 A = R$ are taken.

Would you like a further breakdown of any specific part of this proof? Here are five follow-up questions that could deepen understanding:

1. How do different types of elementary row operations affect the determinant of a matrix?
2. Why does the determinant of an invertible matrix need to be non-zero?
3. Can you explain how the reduced row echelon form $R$ is connected to matrix invertibility?
4. How do you compute the determinant of an elementary matrix?
5. What happens to the determinant of a matrix when a row of zeros appears?

Tip: When computing determinants, remember that row swaps change the sign of the determinant, while multiplying a row by a scalar multiplies the determinant by that scalar.