Question 44: Right-multiplication by an elementary matrix \(E\) affects the columns of A in the same way that left-multiplication affects the rows. Use Theorem 5 and 3 and the obvious fact that \({E^T}\) is another elementary matrix to show that \(\det AE = \left( {\det E} \right)\left( {\det A} \right)\). Do not use Theorem 6.

Theorem 3states that \(A\) be a square matrix in the following conditions:

1. If a multiple on one row of \(A\) is added to another row to produce a matrix \(B\), then \(\det B = \det A\).
2. If two rows of \(A\) are interchanged to produce \(B\), then \(\det B = - \det A\).
3. If one row of \(A\) is multiplied by \(k\) to produce \(B\), then \(\det B = k \cdot \det A\).

Theorem 5states that if \(A\) is an\(n \times n\) matrix, then \(\det {A^T} = \det A\).

According to theorem 5, \(\det AE = {\left( {\det AE} \right)^T}\).

Then\(\det AE = \left( {\det {E^T}{A^T}} \right)\)because \({\left( {AE} \right)^T} = {E^T}{A^T}\). Since \(ET\) is itself an elementary matrix, then according to the proof of theorem 3, \(\det \left( {{E^T}{A^T}} \right) = \left( {\det {E^T}} \right)\left( {\det {A^T}} \right)\). It is true that \(\det AE = \left( {\det {E^T}} \right)\left( {\det {A^T}} \right)\), and by theorem 5, \(\det AE = \left( {\det E} \right)\left( {\det A} \right)\).

Thus, it is proved that \(\det AE = \left( {\det E} \right)\left( {\det A} \right)\).