Finding the determinant of a $4\times4$ matrix using row reduction.

Question

Finding the determinant of a $4\times4$ matrix $$\begin{bmatrix} 1 & 2 & 1 & 0 \\ 0 & 3 & 1 & 1 \\ -1 & 0 & 3 & 1 \\ 3 & 1 & 2 & 0 \end{bmatrix}$$

I did row transformations and finally got upper triangular matrix $$\begin{vmatrix} 1 & 2 & 1 & 0 \\ 0 & 3 & 1 & 1 \\ -1 & 0 & 3 & 1 \\ 3 & 1 & 2 & 0 \end{vmatrix}_{R_3+R_1->R_3\mbox{ and } R_4-3R_1->R_4}$$ $$\begin{vmatrix} 1 & 2 & 1 & 0 \\ 0 & 3 & 1 & 1 \\ 0 & 2 & 4 & 1 \\ 0 & -5 & -1 & 0 \end{vmatrix}_{3R_3-2R_2\mbox{ and }3R_4+5R_2->R_4}$$ $$\begin{vmatrix} 1 & 2 & 1 & 0 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 10 & 1 \\ 0 & 0 & 2 & 5 \end{vmatrix}_{5r_4-R_3->R_4}$$ $$\begin{vmatrix} 1 & 2 & 1 & 0 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 10 & 1 \\ 0 & 0 & 0 & 24 \end{vmatrix}$$ To find the determinant I need to multiply the diagonal elements of the matrix. So, $1\times3\times10\times24=720$. But, however the answer is $16$. I cannot understand where I did a mistake.

Answer 1

As I said in my comment, you can add a multiple of one row to another without changing the determinant, so $R_1+R_3\to R_3$ is fine. However, when you do $3R_4+5R_2\to R_4,$ you have really done two operations: $$\begin{align} 3R_4&\to R_4\\ R_4+5R_2&\to R_4 \end{align}$$ The first of these multiplies the determinant by $3$.

Answer 2

We need also to take into account the coefficients we have used to combine the rows indeed multiplying a row by $k$ of $A$ to obtain $A’$, we have $$\det(A’)=k\cdot \det(A)$$

Answer 3

Let me tell you, that your triangular matrix is wrong. It would have been possible for us to spot out the mistake, if you had given more steps.

Look at the Image below, Thats your triangular matrix(Using Matrix Calculator)

The determinant of the matrix is 16. The Triangular Matrix is

$$ \begin{bmatrix} 1 & 2 &1 &0 \\ 0 & 3 &1 &1 \\ 0 & 0 &\frac{10}{3} &\frac{1}{3} \\0 & 0 &0 &\frac{8}{5} \\ \end{bmatrix}$$