Expansion Theorem for Determinants

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Theorem

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$:


$\ds \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$

where:

the summation $\ds \sum_\lambda$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$
$\map \sgn \lambda$ is the sign of the permutation $\lambda$.


Let $a_{p q}$ be an element of $\mathbf A$.

Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$.


Then:

$(1): \quad \ds \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
$(2): \quad \ds \forall c \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k c} A_{k c}$


Thus the value of a determinant can be found either by:

multiplying all the elements in a row by their cofactors and adding up the products

or:

multiplying all the elements in a column by their cofactors and adding up the products.


The identity:

$\ds D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$

is known as the expansion of $D$ in terms of row $r$, while:

$\ds D = \sum_{k \mathop = 1}^n a_{k c} A_{k c}$

is known as the expansion of $D$ in terms of column $c$.


Corollary

Let $\delta_{rs}$ be the Kronecker delta.

Then:

$(1): \quad \ds \forall r \in \closedint 1 n: \sum_{k \mathop = 1}^n a_{r k} A_{s k} = \delta_{r s} D$
$(2): \quad \ds \forall r \in \closedint 1 n: \sum_{k \mathop = 1}^n a_{k r} A_{k s} = \delta_{r s} D$


That is, if you multiply each element of a row or column by the cofactor of another row or column, the sum of those products is zero.


Proof

Because of Determinant of Transpose, it is necessary to prove only one of these identities.


Let:

$D = \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix}$


First, note that from Determinant with Row Multiplied by Constant, we have:

$\begin{vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r 1} & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end {vmatrix} = a_{r 1} \begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end {vmatrix}$


and similarly:

$\begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & a_{r 2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end {vmatrix} = a_{r 2} \begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{vmatrix}$

and so on for the whole of row $r$.


From Determinant as Sum of Determinants:

$\begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} = \ds \sum_{k \mathop = 1}^n \paren {a_{r k} \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ 0 & \cdots & 1 & \cdots & 0 \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} }$


Consider the determinant:

$\begin{vmatrix} a_{1 1} & \cdots & a_{1 \paren {k - 1} } & a_{1 k} & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} k} & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} k} & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\ \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n k} & a_{n \paren {k + 1} } & \cdots & a_{n n} \end {vmatrix}$


Exchange rows $r$ and $r - 1$, then (the new) row $r - 1$ with row $r - 2$, until finally row $r$ is at the top.

Row $1$ will be in row $2$, row $2$ in row $3$, and so on.

This is permuting the rows by a $k$-cycle of length $r$.

Call that $k$-cycle $\rho$.

Then from Parity of K-Cycle:

$\map \sgn \rho = \paren {-1}^{r - 1}$


Thus:

$\begin {vmatrix} 0 & \cdots & 1 & \cdots & 0 \\ a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} = \paren {-1}^{r - 1} \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\ 0 & \cdots & 1 & \cdots & 0 \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix}$


The same argument can be applied to columns.

Thus:

$\begin {vmatrix} 1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ a_{1 k} & a_{1 1} & \cdots & a_{1 \paren {k - 1} } & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n k} & a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n \paren {k + 1} } & \cdots & a_{n n} \end {vmatrix} = \paren {-1}^{k-1}\begin {vmatrix} 0 & \cdots & 1 & \cdots & 0 \\ a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix}$


and so:

$\begin{vmatrix} 1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ a_{1 k} & a_{1 1} & \cdots & a_{1 \paren {k - 1}} & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n k} & a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n \paren {k + 1} } & \cdots & a_{n n} \end {vmatrix} = \paren {-1}^{r + k} \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\ 0 & \cdots & 1 & \cdots & 0 \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix}$


Then:

$\paren {-1}^{r + k} \begin {vmatrix} a_{1 1} & \cdots & a_{1 \paren {k - 1} } & a_{1 \paren {k + 1} } & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n \paren {k - 1} } & a_{n \paren {k + 1} } & \cdots & a_{n n} \end{vmatrix}$

is $A_{r k}$, the cofactor of $a_{r k}$ in $D$.


But from Determinant with Unit Element in Otherwise Zero Row, we have:

$\begin {vmatrix} 1 & 0 & \cdots & 0 \\ b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix} = \begin {vmatrix} b_{2 2} & \cdots & b_{2 n} \\ \vdots & \ddots & \vdots \\ b_{n 2} & \cdots & b_{n n} \end {vmatrix}$


Assembling all the pieces derived above, the result follows.

$\blacksquare$


Examples

Arbitrary Order $3$ Example

Consider the order $3$ determinant:

$\begin {vmatrix} 1 & 2 & 3 \\ 4 & 1 & 2 \\ 6 & 5 & 4 \end {vmatrix}$

We have:

\(\ds \begin {vmatrix} 1 & 2 & 3 \\ 4 & 1 & 2 \\ 6 & 5 & 4 \end {vmatrix}\) \(=\) \(\ds 1 \times \begin {vmatrix} 1 & 2 \\ 5 & 4 \end {vmatrix} - 2 \times \begin {vmatrix} 4 & 2 \\ 6 & 4 \end {vmatrix} + 3 \times \begin {vmatrix} 4 & 1 \\ 6 & 5 \end {vmatrix}\)
\(\ds \) \(=\) \(\ds 1 \times \paren {1 \times 4 - 2 \times 5} - 2 \times \paren {4 \times 4 - 2 \times 6} + 3 \times \paren {4 \times 1 - 6 \times 5}\)
\(\ds \) \(=\) \(\ds -6 - 8 + 42\)
\(\ds \) \(=\) \(\ds 28\)


Sources

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