2.2. Suppose that each of the row sums of an
n x n matrix A is equal to zero. Show that A
must be singular.
 To show that a matrix $A$ is singular, we need to show that its determinant is equal to 0.

Let $A$ be an $n \times n$ matrix such that each of its row sums is equal to zero. That is, for each row $i$, we have

$$a_{i1} + a_{i2} + \dots + a_{in} = 0.$$

Now, consider the matrix $B$ obtained by subtracting the first row of $A$ from all the other rows. Then, the $(i+1)$th row of $B$ is given by

$$b_{i+1,j} = a_{i+1,j} - a_{1,j}$$

for $i = 1, \dots, n-1$ and $j = 1, \dots, n$. Note that the row sum of each row of $B$ is still equal to zero, since

$$b_{i+1,1} + b_{i+1,2} + \dots + b_{i+1,n} = (a_{i+1,1} - a_{1,1}) + (a_{i+1,2} - a_{1,2}) + \dots + (a_{i+1,n} - a_{1,n}) = a_{
  Suppose that each of the row sums of an $n \times n$ matrix $A$ is equal to zero. We need to show that $A$ must be singular.

Since each row sum of $A$ is zero, we can write:

$$\sum_{i=1}^n a_{i,j} = 0$$

for every $j = 1, 2, \ldots, n$.

Now, let $j \neq k$ be any two integers such that $j, k \in [n]$. Using the above equation, we have:

$$\sum_{i=1}^n a_{i,j} - \sum_{i=1}^n a_{i,k} = 0$$

Expanding the sum and using the fact that the terms are distinct, we get:

$$a_{j,j} - a_{k,k} = 0$$

Since $j \neq k$, we have $a_{j,j} \neq 0$ and $a_{k,k} \neq 0$. Therefore, we have:

$$a_{j,k} = 0$$

This means that the matrix $A$ has a zero row, which means that it is singular.

Therefore, we have shown that if each of the row sums of an $n \times n