##### Proof of TheoremDT

With our definition of the determinant (Definition DM) and theorems like Theorem DER, using induction (Proof Technique I) is a natural approach to proving properties of determinants. And so it is here. Let $n$ be the size of the matrix $A$, and we will use induction on $n$.

For $n=1$, the transpose of a matrix is identical to the original matrix, so vacuously, the determinants are equal.

Now assume the result is true for matrices of size $n-1$. Then, \begin{align*} \detname{\transpose{A}} &=\frac{1}{n}\sum_{i=1}^{n}\detname{\transpose{A}}\\ &= \frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}(-1)^{i+j} \matrixentry{\transpose{A}}{ij}\detname{\submatrix{\transpose{A}}{i}{j}} &&\text{Theorem DER}\\ &= \frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}(-1)^{i+j} \matrixentry{A}{ji}\detname{\submatrix{\transpose{A}}{i}{j}} &&\text{Definition TM}\\ &= \frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}(-1)^{i+j} \matrixentry{A}{ji}\detname{\transpose{\left(\submatrix{A}{j}{i}\right)}} &&\text{Definition TM}\\ &= \frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}(-1)^{i+j} \matrixentry{A}{ji}\detname{\submatrix{A}{j}{i}} &&\text{Induction Hypothesis}\\ &= \frac{1}{n}\sum_{j=1}^{n}\sum_{i=1}^{n}(-1)^{j+i} \matrixentry{A}{ji}\detname{\submatrix{A}{j}{i}} &&\text{Property CACN}\\ &= \frac{1}{n}\sum_{j=1}^{n}\detname{A} &&\text{Theorem DER}\\ &=\detname{A} \end{align*}

Theorem DER⟩  ⟨Definition TM⟩  ⟨Definition TM⟩  ⟨Property CACN⟩  ⟨Theorem DER⟩