# Prove that a determinant is zero without using expansion

How to prove that the following determinant is zero without using expansion?

$$\left|\begin{array}{ccc} 1 & 4 & 1 \\ 2 & -1 & 0 \\ 0 & 18 & 4\end{array}\right|=0$$

I can't get any 2 rows or any 2 columns to be equal and I can't get an entire row or entire column to be zero? What series of operations are required? Thanks.

Edit

I noticed that it is easy to get the diagonal to be zero. I am not sure when one could say if the diagonal is zero then the determinant is zero...

These will do the job for us: $$R_3 \to R_3 - 4R_1$$ $$R_3 \to R_3 + 2R_2$$

• Nevermind dude! Mar 6, 2017 at 11:15

Four times the first row is $(4,16,4)$.

$-2$ times the second row is $(-4,2,0)$.

Adding these up gives the third row $(0,18,4)$.

Hence, the rows of the given matrix have the relation $4R_1 -2R_2 - R_3 = 0$, hence it follows that the determinant of the matrix is zero as the matrix is not full rank.

EDIT : The rank of a matrix, is the dimension of it's image as a linear operator. That is, when we treat the rows (or columns) of the matrix as a vector and take their span, the dimension of that vector space is called the rank.

If a matrix has the maximum rank possible (which is the dimension of the matrix), then it is said to be of full rank.

If the matrix is not of full rank, then say you can write $R_1 = \sum_{i=2}^n c_i R_i$, because of linear dependence. Then, just do the row operations $R_1 \to R_1 - c_iR_i$ for each $i=2$ to $n$. This will keep the determinant unchanged, and make the row $R_1$ as zero, so the determinant of the matrix will be zero.

• Could you please elaborate on the meaning of full rank? Mar 6, 2017 at 10:51
• Done, answer edited. Mar 6, 2017 at 11:02
• Is there a mathematical approach to find the values 4, -2, -1 that form the equation 4R1-2R2-R3=0? I mean, given an equation such as aR1+bR2+cR3=0, can we find a, b and c using a known convergent process without using brute-force? I could start another question if you think it is more appropriate to do so. Mar 7, 2017 at 11:03
• Well, I got this one figured on my own. No need to trouble you. Thx, Mar 7, 2017 at 21:29
• It's good you did, unfortunately I was out the last few days. Mar 8, 2017 at 23:08