## Matrices containing a row (or column) of 0's

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 $$\det(A) = 0$$.

## Permutation matrices

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 $$\sigma$$.

## Triangular matrices

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.

## Exercises

Compute the determinants of each of the following matrices:

1. $$\begin{bmatrix} 2 & 3 \\ 0 & 2\end{bmatrix}$$

2. $$\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}$$

3. $$\begin{bmatrix} 2-i & 0 \\ 3 & 1+i\end{bmatrix}$$