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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.

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

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

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

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