Finding determinant for matrix using upper triangle method

Question

Here is an example of a matrix, and I'm trying to evaluate its determinant:

$$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 2 & 5 & -2 & 9 \\ 3 & 7 & 0 & 1 \\ \end{pmatrix} $$

When applying first row operation i get:

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

Now, if I continue doing row operations until i get the upper triangle, the determinant will be 14 (which is said to be the correct one).

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

However, if I instead apply this certain operation, R4 --> (1/-2)R4...

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

...and then carry on with operations, I get a different final answer: The determinant will be 7 in this case!

$$ \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 1 & 4 & -4 \\ 0 & 0 & -1 & 5 \\ 0 & 0 & 0 & -7 \\ \end{pmatrix} $$

Could someone explain that to me - is this operation illegal? R4 --> (1/-2)R4

How so? Because i always tend to use it, just to simply things a little.

Answer 1

You just multiplied a row with $\frac {1}{-2}$! This will change the value of determinant. What you can do is take $-2$ common from a row and write it outside.

Consider a $1\times 1$ matrix $A=[1]$.

$det(A)=1$

Apply $R_!\to2R_1$

$A=[2]$

$det(A)=2$

Can you see why you cannot do it?

Answer 2

Different row-operations affect the determinant of the matrix differently. Adding a multiple of one row to another will not change the determinant. However, multiplying a row by some factor will lead to the determinant being multiplied by the same factor.

So, since you multiplied $R_4$ by the factor $- \frac 12$, the resulting determinant will be $- \frac 12$ times what the determinant of the original matrix was.

Answer 3

If we multiply a certain row or column of a matrix $A$ by some scalar $\lambda$ then determinant of $A$ changes to $\lambda|A|.$