[Math] Evaluate det(A) by cofactor expansion along a row or column of your choice. (Smart choice of row or column)

linear algebramatrices

I understand cofactor expansion along a row or column, however I'm confused as to what the steps are once you choose the row or column based on the number of zeros they contain. The first step I understand, in a 4×4 matrix choose the row or column with the most zeros. However, after that I'm lost. In this case I would choose the 3rd column, but which row do I choose after this? And once I get to a 3×3 matrix the same problem occurs. Which row and column do I choose? (ANS is -240)
$$
\begin{bmatrix}
3 & 3 & 0 & 5 \\
2 & 2 & 0 & -2 \\
4 & 1 & -3 & 0 \\
2 & 10 & 3 & 2
\end{bmatrix}
$$

Best Answer

You don't choose a column AND a row. You only choose one column OR one row in your first step, and proceed through that column or row, one element at a time.

You chose the third column. Good choice. Now we move through that column one element at a time. The formula in this case will give us this:

$\left|\begin{matrix} 3 & 3 & 0 & 5 \\ 2 & 2 & 0 & -2 \\ 4 & 1 & -3 & 0 \\ 2 & 10 & 3 & 2 \end{matrix} \right| = 0 \left|\begin{matrix} 2 & 2 & -2 \\ 4 & 1 & 0 \\ 2 & 10 & 2 \end{matrix} \right| - 0\left|\begin{matrix} 3 & 3 & 5 \\ 4 & 1 & 0 \\ 2 & 10 & 2 \end{matrix} \right| - 3\left|\begin{matrix} 3 & 3 & 5 \\ 2 & 2 & -2 \\ 2 & 10 & 2 \end{matrix} \right| - 3\left|\begin{matrix} 3 & 3 & 5 \\ 2 & 2 & -2 \\ 4 & 1 & 0 \end{matrix} \right|$

The first matrix on the RHS was obtained by removing row 1 and column 3 from the original matrix. We do this because that $0$ is in row 1 and column 3.

The second matrix on the RHS was obtained by removing row 2 and column 3 from the original matrix. We do this because that $0$ is in row 2 and column 3. Note that the $0$ has a minus sign in front of it. This is because of the alternating $+-+-$ pattern when doing cofactor expansion determinants.

The third matrix on the RHS was obtained by removing row 3 and column 3 from the original matrix. We do this because that $-3$ is in row 3 and column 3.

The fourth matrix on the RHS was obtained by removing row 4 and column 3 from the original matrix. We do this because that $3$ is in row 4 and column 3. Note that the $3$ has a minus sign in front of it. This is because of the alternating $+-+-$ pattern when doing cofactor expansion determinants.

Let me know if you need more info.

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