# Showing determinant is 0 (given an arbitrary matrix) [closed]

How do I show $\det(A) = 0$ ? $$\det(A) = \begin{vmatrix} 0&0&0&a&b \\ 0&0&0&c&d \\0&0&0&e&f \\ p&q&r&s&t \\ v&w&x&y&z \end{vmatrix}$$

I know that I need to make a row with zero entries which automatically concludes $det(A)=0$ but I don't know how to start. Roughly I did:

$$\begin{vmatrix} 0&0&0&a-c&b-d \\ 0&0&0&c-a&d-b \\ 0&0&0&e&f \\ p&q&r&s&t \\ v&w&x&y&z \end{vmatrix}$$

and from here I don't know how to factor it out. Any help is appreciated!

## closed as off-topic by Namaste, José Carlos Santos, Aweygan, man and laptop, Ben ShellerFeb 8 '18 at 0:01

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• Hint: The first three rows must be linearly dependent. – Trevor Norton Dec 7 '17 at 16:46
• As an aside, in your attempt, you appeared to take row one and subtract row two at the same time as taking row two and subtracting row one. This is not allowed. Such elementary row operations should occur in sequence. By the time you went to have row two minus row one, row one had changed and no longer ended with $a,b$ instead ending with $a-c,b-d$. – JMoravitz Dec 7 '17 at 16:48
• You matrix is a sum of two row vectors and two column vectors (i.e. a sum of 4 matrices with rank $1$). Since rank of a matrix is subadditive (i.e. ${\rm rank}(X+Y) \le {\rm rank}(X)+{\rm rank}(Y))$, the rank of your matrix is at most $4$. Since this is smaller than $5$, the dimension of your matrix. $A$ isn't full rank and $\det(A) = 0$. – achille hui Dec 7 '17 at 16:51
• @JMoravitz thank you for pointing that out! – dembrownies Dec 7 '17 at 17:20
• @achillehui I never see it that way. Thank you! – dembrownies Dec 7 '17 at 17:21

Since the first three rows (as the first three columns) are certainly linearly dependent $rank(A)<5$ and $det(A)=0$.
$$R = \begin{bmatrix} 0&0&0&a&b \\ 0&0&0&c&d \\0&0&0&e&f \end{bmatrix}$$
$$R^* = \begin{bmatrix} 0&0&0&1&0 \\ 0&0&0&0&1 \\0&0&0&0&0 \end{bmatrix}$$