Miscalculating the determinant

Question

I am learning linear algebra and am getting stuck when trying to calculate the determinant using elementary row operations. Consider the matrix A.

\begin{vmatrix} 0 & 1 & 2 & 3 \\ 1 & 1 & 1 & 1 \\ -2 & -2 & 3 & 3 \\ 1 & 2 & -2 & -3 \\ \end{vmatrix}

According to the solution in my textbook and Matlab the determinant should be 10. I however find -20. Here is what I did. I first interchanged row 1 and row 3. \begin{vmatrix} 1 & 2 & -2 & -3 \\ 1 & 1 & 1 & 1 \\ -2 & -2 & 3 & 3 \\ 0 & 1 & 2 & 3 \end{vmatrix}

I then substracted row 1 from row two. I also added the first row twice to the third row. \begin{vmatrix} 1 & 2 & -2 & -3 \\ 0 & -1 & 3 & -2 \\ 0 & 2 & -1 & 9 \\ 0 & 1 & 2 & 3 \end{vmatrix}

Then, I added the second row twice to the third row and once to the fourth row. \begin{vmatrix} 1 & 2 & -2 & -3 \\ 0 & -1 & 3 & -2 \\ 0 & 0 & 5 & 5 \\ 0 & 0 & 5 & 1 \end{vmatrix} My final operation was to substract the third row from the fourth row, which gave: \begin{vmatrix} 1 & 2 & -2 & -3 \\ 0 & -1 & 3 & -2 \\ 0 & 0 & 5 & 5 \\ 0 & 0 & 0 & -4 \end{vmatrix}

Finally, I calculated the determinant: $(-1)^1 \cdot 1 \cdot -1 \cdot 5 \cdot -4 = -20$ The $(-1)^1$ is there since I did one operation in which I interchanged two rows. I would really appreciate if you could tell me what I did wrong.

Martijn

Answer 1

You have a few mistakes in calculations. In step $2$ when you subtract row $1$ from row $2$ it should be $1-(-3)=4$ in the fourth column. In the same step when you add row $1$ twice to row $3$ it should be $3+2\times(-3)=-3$ in the fourth column.

Answer 2

Actually, the first thing that you did was to exchange rows $1$ and $4$, not $1$ and $3$. After that, when you subtracted row $1$ from row $2$, you should have got a $4$ at the end of the row. And when you add twice the first row to the third one, the last entry should become $-3$, not $9$.