$$

Matrices with dependent rows

Problem

Let

$A$ be an

$n\times n$ matrix. Show the following.

1. Suppose
   $A$ contains a zero row. Then
   $det(A)=0$.
2. Suppose
   $A$ contains two identical rows. Then
   $det(A)=0$.
3. Suppose
   $A$ contains a set of rows that is linearly dependent. Then
   $det(A)=0$.

Thought

The key idea here is to show these properties by definition of the determinant–-the four rules. Also, don't try to come up with a proof directly–-look at some examples first!

For Statement 1, observe that none of the four rules in the definition say directly a matrix with a zero row has its determinant zero; that is why we need to prove it. Let's look at an example when

$$A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 1& 2& 4\end{array}\right].$$

The interesting thing about this matrix is that

$A\stackrel{{\rho }_2:\times 2}{\to }A$. By definition, this means

$det(A)\times 2=det(A)$, which can only happens when

$det(A)=0$. Though this argument seems a bit tricky, but it does tell you that

$det(A)=0$.

For Statement 2, think about

$$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 1& 2& 4\end{array}\right].$$

Note that

$A\stackrel{{\rho }_2\leftrightarrow {\rho }_3}{\to }A$.

For Statement 3, think about

$$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 4\\ 2& 3& 5\end{array}\right].$$

Note that

$A\stackrel{{\rho }_3:-{\rho }_1}{\to }\stackrel{{\rho }_3:-{\rho }_2}{\to }B$, where

$B$ is a matrix whose

$3$-rd row is zero.

Sample answer

1. Let
   $A$ be a matrix whose
   $i$-th row is zero. Then we have
   $A\stackrel{{\rho }_i:\times 2}{\to }A$. By definition,
   $det(A)\times 2=det(A)$, which means
   $det(A)=0$.
2. Let
   $A$ be a matrix whose
   $i$-th row and
   $j$-th row are identical^[1]. Then we have
   $A\stackrel{{\rho }_i\leftrightarrow {\rho }_j}{\to }A$. By definition,
   $det(A)\times (-1)=det(A)$, which means
   $det(A)=0$.
3. Let
   $A$ contain a set of rows that is linearly dependent^[2]. Thus, there is a row that can be written as a linear combination of other rows. Let
   ${\mathbf{r}}_1,\dots ,{\mathbf{r}}_n$ be the rows of
   $A$. Without loss of generality, we may assume
   ${\rho }_1=c_2{\mathbf{r}}_2+c_3{\mathbf{r}}_3+\cdots +c_n{\mathbf{r}}_n$. Thus, we have
   $$A\stackrel{{\rho }_1:-c_2{\mathbf{r}}_2}{\to }\stackrel{{\rho }_1:-c_3{\mathbf{r}}_3}{\to }\cdots \stackrel{{\rho }_1:-c_n{\mathbf{r}}_n}{\to }B$$ such that
   $B$ is a matrix whose first row is zero, which means
   $det(B)=0$ by Statement 1. Since row combination does not change the determinant,
   $det(A)=det(B)=0$.

This note can be found at Course website > Learning resources.

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1. In mathematics, the word "identical" means "exactly the same". For example of vectors, it means two vectors have the same length and the corresponding entries are the same. ↩︎

2. Review articles 16 ~ 19 of Learning resources from Linear Algebra I. ↩︎