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# 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

$$
A = \begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 0 \\
1 & 2 & 4
\end{bmatrix}.
$$

The interesting thing about this matrix is that $A \xrightarrow{\rho_2:\times 2} A$. 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

$$
A = \begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 2 & 4
\end{bmatrix}.
$$

Note that $A \xrightarrow{\rho_2\leftrightarrow\rho_3} A$.

For Statement 3, think about

$$
A = \begin{bmatrix}
1 & 1 & 1 \\
1 & 2 & 4 \\
2 & 3 & 5
\end{bmatrix}.
$$

Note that $A \xrightarrow{\rho_3:-\rho_1}\xrightarrow{\rho_3:-\rho_2} B$, 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 $A\xrightarrow{\rho_i:\times 2}A$. 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[^identical]. Then we have $A\xrightarrow{\rho_i\leftrightarrow\rho_j}A$. 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[^dependent]. Thus, there is a row that can be written as a linear combination of other rows. Let $\br_1, \ldots, \br_n$ be the rows of $A$. Without loss of generality, we may assume $\rho_1 = c_2\br_2 + c_3\br_3 + \cdots + c_n\br_n$. Thus, we have
$$
A \xrightarrow{\rho_1:-c_2\br_2} \xrightarrow{\rho_1:-c_3\br_3} \cdots \xrightarrow{\rho_1:-c_n\br_n} B
$$
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, $\det(A) = \det(B) = 0$.

[^identical]: 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.

[^dependent]: Review articles 16 ~ 19 of [Learning resources](https://hackmd.io/@jephianlin/2023FMath103A-resources) from Linear Algebra I.

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