Justify, without evaluating, that the determinant of the following matrix is zero

I am currently stuck at this question and have no idea how to solve. I just started out learning linear and I'm really weak in this field.

Justify, without evaluating, that the determinant of the following matrix is zero

Here's the matrix A:

$$\begin{pmatrix} 1 & 0 & 2 & 4\\ -2 & 3 & 8 & 6\\ -1 & 3 & 10 & 10\\ 6 & 6 & -3 & 7\\ \end{pmatrix}$$

I searched online but couldn't find something similar. What I found though was that if it was skew-symmetric ($A^t= -A$) then the determinant could directly be said to be equal to zero. But in this case it didn't work with me.

Thank you.

The third row is the sum of the first and second rows. The rows are not linearly independent, so the determinant is zero.

• Does the same logic apply to columns as well? – Kode Ch Feb 25 '18 at 13:54
• @KodeCh Yes, it does, see this answer – A. Goodier Feb 25 '18 at 13:56
• Thank you for your help – Kode Ch Feb 25 '18 at 13:58

The easiest solution is by far the one from @A.Goodier.

Still, if you're stuck, you could try to solve

$$\left\{ \begin{array}{c} a &+ 0b &+ 2c &+ 4d&=0 \\ -2a &+ 3b&+8c&+6d&=0 \\ -a&+ 3b&+10c&+10d&=0 \\ 6a&+ 6b&-3c&+7d&=0 \end{array} \right.$$

which is the equivalent of trying to find the vectors $V$ for which $M * V = 0$.

If you find a solution $V$ which isn't the zero vector, it means the columns are linearly dependent and the determinant of the matrix is $0$.

Solving the equation, you find that:

$$\left\{ \begin{array}{c} b &= \frac{a}{33} \\ c &= \frac{15 a}{22} \\ d &= -\frac{13 a}{22}\\ \end{array} \right.$$

For example, with $a = 66$, you find

$$V= \begin{bmatrix} 66 \\ 2 \\ 45 \\ -39 \end{bmatrix}$$