Justify that the determinant of the following matrix is zero $$A=\begin{pmatrix} 2 & 1 & 0 & 5\\ -1 & 1 & 1 & 6\\ 5 & 1 & -1 & 4\\ 5 & 1 & 3 & 0 \end{pmatrix}$$
I have two questions about this example task:
If a task says "justify / reason why", am I allowed to do any calculation or am I only allowed to write text?
Do you have a better way than mine and can you please show it (I needed to use Gaussian elimination which was very exhausting)?
The determinant of a square matrix is zero if its row(-vectors) are linearly dependent. Its row(-vectors) are linearly dependent if we get a row with only zeroes and for this we use Gaussian elimination.
Now just apply Gaussian elimination and get a row with only zeroes (I won't do it here to keep it short so just assume we ended up like that).
Would this be a correct way of reasoning or are calculations entirely forbidden for such worded tasks in general? :s
If you find that some row or colum is a linear combination of the others then you can conlude without further calculation otherwise you need to evaluate the determinant or perform RREF.
Sometimes we can also use mixed methods by partial RREF and calculation of the determinant by Laplace method.
In this case we can see that $-2R1+R2+R3=0$ then $$\begin{pmatrix} 2 & 1 & 0 & 5\\ -1 & 1 & 1 & 6\\ 5 & 1 & -1 & 4\\5&1&3&0 \end{pmatrix}\to \begin{pmatrix} 2 & 1 & 0 & 5\\ -1 & 1 & 1 & 6\\ 0 & 0 & 0 & 0\\5&1&3&0 \end{pmatrix}$$
and we can conclude easily that $\det=0$.