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Linear Algebra

Formulae for the determinant

Previously we gave a specification for how the determinant should behave. We use this to deduce some rather impressive features of the determinant. Note that we can say the determinant as we know any two determinants are equal.

Here we turn to giving one determinant and a formula. The formula is, perhaps, the most common one. However, computing determinants through Gaussian Elimination is generally faster. The Bareiss algorithm is the most common one used.

We need some notation to concisely state the formula here. For a $m \times n$ matrix $A$, the $(i,j)$ minor of $A$ is the $(m-1) \times (n-1)$ matrix obtained by deleting the i-th row and j-th column of $A$. We denote the $(i,j)$-minor of $A$ by $M_{ij}(A)$ or, if the context allows, just by $M_{ij}$.

For example, if \(A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}\)

Then, we have \(M_{11} = \begin{pmatrix} 5 & 6 \\ 8 & 9 \end{pmatrix}, \ M_{21} = \begin{pmatrix} 2 & 3 \\ 8 & 9 \end{pmatrix}, \ M_{23} = \begin{pmatrix} 1 & 2 \\ 7 & 8 \end{pmatrix}\)

Definition. We define a function \(\det : \operatorname{Mat}_{n \times n}(k) \to k\) recursively on $n$. For $n=1$, we set \(\det(a) := a\) For $n > 1$, we set \(\det A := \sum_{i=1}^n (-1)^{i-1} A_{i1} \det M_{i 1}\)

This formula is often called a cofactor expansion for the determinant.

Let’s write out a closed form formula for small values of $n$:

  • Since $n=1$ is already given, let’s take a look at $n=2$. Then, for \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) we have \(\det A = a \det(d) - c \det(b) = ad-bc\)
  • For $n=3$, with \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) we have \(\begin{aligned} \det A & = a \det \begin{pmatrix} e & f \\ h & i \end{pmatrix} - d \det \begin{pmatrix} b & c \\ h & i \end{pmatrix} + g \det \begin{pmatrix} b & c \\ e & f \end{pmatrix} \\ & = a (ei -fh) - d(bi-ch) + g(bf-ce) \\ & = aei - afh - bdi + bfg + cdh - ceg \end{aligned}\)

At this point, you realize you never want to try to remember the closed form for the cofactor expansion. The recursive definition is far more amenable to human consumption and usage.

Now, let’s state that this definition actually meets the specification for a determinant.

A proof is not necessarily hard, in terms of requiring new ideas, but it is too time consuming for us. We will therefore content ourselves with a skim of the ideas involved.

  • First, one shows that \(\det (P(i,j) A) = - \det A\) This comes by working inductively and carefully analyzing the terms in the sum \(\det (P(i,j) A) = \sum_{l=1}^n (-1)^{l-1} (P(i,j)A)_{l1} \det(M_{l1}(P(i,j)A))\) and comparing them to $A_{l1}$ and $P(u,v)M_{l1}(A)$.

  • Next, one checks that \(\det (S(i,c) A) = c \det A\) This one is a bit simpler but follows the same lines: working via induction and analyzing the terms in the sum for the cofactor expansion.

  • Finally, one checks that \(\det (L(i,j,c)A) = \det A\) again working along the same lines.

We see that $\det$ is multiplicative when using the elementary operation matrices on $A$. It turns out this enough.

In another direction, it turns out the multiplicativity of a determinant is actually determined by multiplicativity using elementary row operations.

Theorem. Assume \(\operatorname{det}^\prime : \operatorname{Mat}_{n \times n}(k) \to k\) satisfying

  • \[\operatorname{det}^\prime(L(i,j,c) A) = \operatorname{det}^\prime A\]
  • \[\operatorname{det}^\prime(S(i,c)A) = c \operatorname{det}^\prime A\]
  • \[\operatorname{det}^\prime(P(i,j) A) = - \operatorname{det}^\prime A\]
  • \[\operatorname{det}^\prime(I_n) = 1\]

Then, $\det^\prime$ is a determinant.

Corollary. The cofactor expansion formula gives a determinant.

In fact, one can prove the more general cofactor expansion formula for the determinant.

Proposition. For any $1 \leq j \leq n$, we have \(\det A = \sum_{i = 1}^{n=1} (-1)^{i+j-1} A_{ij} \det M_{ij}\)

This allows us to expand along any column at the cost of a sign.

We can also expand along rows.

Proposition. For any $1 \leq i \leq n$, we have \(\det A = \sum_{j = 1}^{n=1} (-1)^{i+j-1} A_{ij} \det M_{ij}\)

For a fuller discussion of the details of the results, a good reference is the suggested textbook Chapter 4 Section I.4 of Hefferon’s Linear Algebra.

For us, it is important to know that

  • determinants exist and
  • there are many interesting formulas to compute them.

But, we understand determinants through their properties and compute them using Gaussian Elimination or a factorization.