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

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


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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Linear Algebra

Inverse

Matrices

Determinant

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