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
Best Answer
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.