Let \(A\) be a square matrix with a row or a column of 0's. Then \(\det(A) = 0\). To see that, notice that every term in the definition of \(\det(A)\) is a product of the form \(A_{1,\sigma(1)} A_{2,\sigma(2)} \cdots A_{n,\sigma(n)}\) for some permutation \(\sigma\). Hence, each term contains exactly one entry from each row and each column of \(A\), implying that every term is 0. So \(\det(A) = 0\).

An \(n\times n\) permutation matrix is a matrix obtained from the \(n\times n\) identity matrix by permuting its rows. For example, \(\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \end{bmatrix}\) is a permutation matrix.

As the name suggests, an \(n\times n\) permutation matrix provides an encoding of a permutation of the set \(\{1,\ldots,n\}\). One interpretation is as follows: If \(\sigma\) is the permutation the matrix encodes, then \(\sigma(i)\) is given by the column index of the entry containing the \(1\) in row \(i\). For the example above, the permutation would be \(\left(\begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{array} \right)\) because in row 1, column 2 contains 1; in row 2, column 4 contains 1; in row 3, column 1 contains 1; in row 4, column 3 contains 1.

Given an \(n\times n\) permutation matrix \(P\) encoding the permutation \(\sigma\), the determinant of \(P\) is simply \((-1)^{\text{#inv}(\sigma)}\). For the example above, there are three inversions. So the determinant is \((-1)^3 = -1\).

This can be readily seen from the definition of the determinant: As each term in the definition consists of \((-1)^{\text{#inv}(\sigma')}\) for some permutation \(\sigma'\) times the product of \(n\) entries from the matrix, exactly one from each row and one from each column, the only way we get a nonzero term from \(P\) is to have a permutation that picks the 1 from each row. The only permutation that does that is \(\sigma\).

Let \(A\) be an upper triangular square matrix. That is, \(A_{i,j} = 0\) whenever \(i \gt j\). For example, the matrix \(\begin{bmatrix} \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{0}\\ 0 & \mathbf{2} & \mathbf{5} & \mathbf{6}\\ 0 & 0 & \mathbf{3} & \mathbf{7}\\ 0 & 0 & 0 & \mathbf{1} \end{bmatrix}\) is upper triangular.

Then **\(\det(A)\) is given by the product
of the diagonal entries**. For the example above,
the determinant is \(1\cdot 2\cdot 3\cdot 1 = 6\).

Let us see why this is the case. Let \(\sigma \in S_n\). We claim that if \(\sigma\) is not the identity permutation, then \(\displaystyle\prod_{i = 1}^n A_{i, \sigma(i)} = 0\).

Suppose that \(\sigma(1) \neq 1\). Then there must be some \(i \geq 2\) such that \(\sigma(i) = 1\). This gives \(A_{i,\sigma(i)} = 0\) since \(A\) is upper triangular and \(i > \sigma(i)\). Hence, \(\displaystyle\prod_{i = 1}^n A_{i, \sigma(i)} = 0\) if \(\sigma(1) \neq 1\).

So suppose that \(\sigma(1) = 1\) but \(\sigma(2) \neq 2\). Then there is some \(i \neq 2\) such that \(\sigma(i) = 2\). But \(i\neq 1\) since we already have \(\sigma(1) = 1\). Hence, \(i \geq 3\). This again gives, \(A_{i,\sigma(i)} = 0\) since \(i > \sigma(i)\). Hence, \(\displaystyle\prod_{i = 1}^n A_{i, \sigma(i)} = 0\) for such a \(\sigma\).

One can continue in this fashion to show that if \(\sigma\) is such that \(\sigma(i) = i\) and \(\sigma(i+1)\neq i+1\), then \(\displaystyle\prod_{i = 1}^n A_{i, \sigma(i)} = 0\). Hence, the only term in \(\det(A)\) that can be nonzero is when \(\sigma(i) = i\) for all \(i=1,\ldots,n\), implying that \(\det(A) = A_{1,1}A_{2,2}\cdots A_{n,n}\).

Using a similar argument, one can conclude that the determinant of a lower triangular matrix (a matrix in which all the entries above the diagonal are 0) is given by the product of the diagonal entries as well.

Compute the determinants of each of the following matrices:

\(\begin{bmatrix} 2 & 3 \\ 0 & 2\end{bmatrix}\)

\(\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}\)

\(\begin{bmatrix} 2-i & 0 \\ 3 & 1+i\end{bmatrix}\)