Troubles on a determinat of a $4\times 4$ matrix

Question

$$A = \left( \begin{array}{cccc} 1 & 3 & 1 & 0 \\ 1 & -1 & 2 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 2 \\ \end{array} \right)$$

Its determinant is $7$ for I used Laplace method. I checked and it's $7$.

When I compute Guass reduction, though it is not $7$ anymore. I took into account the scale factor, but I do not know where I did wrong.

Those the operations I've made: the first two ones:

$$\text{Row 2} - \text{row 1} \to \text{row 2}$$ $$\text{row 3} - \text{row 1} \to \text{row 3}$$

No scale factor.

Then

$$\text{4 row 3} - \text{3 row 2} \to \text{row 3}$$ $$\text{3 row 4} - \text{row 3} \to \text{row 4}$$

At this point the matrix is

$$A = \left( \begin{array}{cccc} 1 & 3 & 1 & 0 \\ 0 & -4 & 1 & 1 \\ 0 & 0 & -3 & 1 \\ 0 & 0 & 0 & 7 \\ \end{array} \right)$$

Dividing now the last row by $7$ and I get

$$A = \left( \begin{array}{cccc} 1 & 3 & 1 & 0 \\ 0 & -4 & 1 & 1 \\ 0 & 0 & -3 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$

Now the scale factor is $7$.

Yet this matrix is a scale matrix, hence the determinant is trivial that is $1\cdot (-4)\cdot (-3)\cdot 1 = 12$

And $12\cdot 7 \neq 7$. Where have I went wrong?

Answer 1

When yu add or subtract a multiple of one row to/from another row, the coefficient of the latter row should be $1$. Thus properly:

$$\text{Row 2} - \text{row 1} \to \text{row 2}$$ $$\text{row 3} - \text{row 1} \to \text{row 3}$$

as given in the question, but then

Answer 2

When you add or subtract a multiple of one row to/from another row, the coefficient of the latter row should be $1$. Thus properly:

$$\text{Row 2} - \text{row 1} \to \text{row 2}$$ $$\text{row 3} - \text{row 1} \to \text{row 3}$$

as given in the question, but then

$$\text{row 3} - \text{(3/4) row 2} \to \text{row 3}$$ $$\text{row 4} - \text{(1/3) row 3} \to \text{row 4}$$

where we divided by $3$ or $4$ to keep the target rows on the same scale.

Answer 3

While computing the determinant of a square matrix using the elementary row operations of the form $R_i=aR_i+bR_j$; just take care that either you choose $a=1$ strictly or if not, then divide the matrix by factor $1/a$. Since you have already used two such operations going with the later choice so you need to balance the determinant value of this matrix by $\frac{1}{4}.\frac{1}{3}=\frac{1}{12}.$