A determinant can be expanded along any row or any column, not only the first: multiply each entry of that line by its cofactor and add. The skill is in choosing the line with the most zeros, since a zero entry takes its whole cofactor out of the work.

Counting the people in a hall row by row or column by column gives the same total, and it is quickest to begin where most seats are empty.

For an $n\times n$ matrix and any fixed row $i$, the determinant is $a_{i1}C_{i1}+\cdots+a_{in}C_{in}$. For any fixed column $j$ it is $a_{1j}C_{1j}+\cdots+a_{nj}C_{nj}$. This is Laplace's expansion. Each cofactor is a determinant one size smaller, so an $n\times n$ determinant becomes a combination of $(n-1)\times(n-1)$ ones, and so on down to $2\times2$. That the first row works for every $n$ is the grouping argument of the last lesson with more terms; its sign bookkeeping is taken on trust here.

• Expand along the row or column with the most zeros: every zero entry removes its cofactor from the work.
• Create zeros first. Adding multiples of one row to another keeps the determinant and can empty most of a column.
• Columns work as well as rows, $\det A=\sum_{i}a_{ij}C_{ij}$ for each column , because .

Move row $i$ to the top by exchanging it with each row above it in turn. That is $i-1$ exchanges, so the determinant changes sign $i-1$ times, while the other rows keep their order, so every minor is unchanged. Expanding the new matrix along its first row and restoring the sign gives the formula.

Exchange row $i$ with the row above it, then with the one above that, until it is first. The other rows keep their order.

That takes $i-1$ exchanges, and each exchange reverses the sign.

Deleting the first row of $B$ and column $j$ leaves exactly what deleting row $i$ and column $j$ of $A$ leaves, in the same order.

Expand $B$ along its first row, which holds the entries of row $i$ of $A$.

Multiply by $(-1)^{i-1}$: the signs combine into $(-1)^{i+j}$, and each term is an entry times its cofactor.

Count the zeros in each row and each column. A zero entry contributes nothing, so its minor never has to be computed.

Column 2 has three zeros. Expanding along it leaves one term, $a_{32}C_{32}=4\cdot(-M_{32})$, and , so .

• $\det A$ can be expanded along any row or any column: entries times cofactors, added.
• Choose the line with the most zeros, and create zeros with row operations first.
• The signs follow the chessboard pattern $(-1)^{i+j}$.
• Entries of one row times the cofactors of another row add up to $0$.

Great work! You now know how to:

• expand a determinant along any row or column;
• choose the line that saves the most work;
• combine row operations with expansion;
• explain why every row gives the same value.