KindiakMath

What is a Determinant?

Many courses in linear algebra discuss determinants, since such a discussion empowers us to compute eigenvalues, which finds its way into many STEM applications. Yet, the theory of determinants is rather challenging to set up in a rigorous manner. To motivate our discussion, we will analyse the 2 \times 2 case.

Let \mathbf A := \begin{bmatrix}a & b \\ c & d\end{bmatrix} \in \mathcal M_{2 \times 2}(\mathbb K) denote a 2 \times 2 matrix.

Lemma 1. \mathbf A is injective if and only if ad-bc \neq 0.

Proof. Now if a = c = 0, then \mathbf A has a zero column and thus is not injective. Furthermore, ad - bc = 0 - 0 = 0.

Therefore, suppose a \neq 0 without loss of generality. We make the row operations

\mathbf A = \begin{bmatrix}a & b \\ c & d\end{bmatrix} \longrightarrow \begin{bmatrix}a & b \\ ac & ad\end{bmatrix} \longrightarrow \begin{bmatrix}a & b \\ 0 & ad-bc\end{bmatrix} =: \mathbf R.

Recall that \mathbf A is injective if and only if \mathbf R is, which holds if and only if ad - bc \neq 0.

The quantity ad - bc is precisely the determinant of a 2 \times 2 matrix.

Theorem 1. The determinant of \mathbf A is defined by

\det(\mathbf A) \equiv \left| \begin{matrix}a & b \\ c & d\end{matrix} \right| := ad - bc.

Consequently, \mathbf A is invertible if and only if \det(\mathbf A) \neq 0.

Proof. If \det(\mathbf A) \neq 0, then \mathbf A is injective. By the fundamental theorem of invertible matrices, \mathbf A is bijective.

The determinant, therefore, tells us whether a matrix is invertible or not. This still raises a challenging question: in the case \mathbf A^{-1} actually exists, how do we compute it?

Theorem 2. If \mathbf A is invertible, then

\displaystyle \begin{aligned}\mathbf A^{-1} &= \frac{1}{\det(\mathbf A)} \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}.\end{aligned}

Proof. We have

\mathbf A\mathbf e_1 = a \mathbf e_1 + c \mathbf e_2,\quad \mathbf A\mathbf e_2 = b \mathbf e_1 + d \mathbf e_2.

Taking inverses,

\mathbf e_1 = a \mathbf A^{-1} \mathbf e_1 + c \mathbf A^{-1} \mathbf e_2,\quad \mathbf e_2 = b \mathbf A^{-1} \mathbf e_1 + d \mathbf A^{-1} \mathbf e_2.

Multiplying by relevant constants,

d\mathbf e_1 = ad \mathbf A^{-1} \mathbf e_1 + cd \mathbf A^{-1} \mathbf e_2,\quad c\mathbf e_2 = bc \mathbf A^{-1} \mathbf e_1 + cd \mathbf A^{-1} \mathbf e_2.

Subtracting the equations,

(ad-bc) \mathbf A^{-1} \mathbf e_1 = d\mathbf e_1 - c\mathbf e_2 = \begin{bmatrix} d \\ -c\end{bmatrix}.

Hence,

\displaystyle \mathbf A^{-1} \mathbf e_1 = \frac{1}{ad - bc} (d\mathbf e_1 - c\mathbf e_2) = \frac{1}{\det(\mathbf A)} \begin{bmatrix} d \\ -c\end{bmatrix}.

A similar computation for \mathbf A^{-1} \mathbf e_2 yields

\displaystyle \mathbf A^{-1} \mathbf e_2 = \frac{1}{ad - bc} (-b\mathbf e_1 + a\mathbf e_2) = \frac{1}{\det(\mathbf A)} \begin{bmatrix} -b \\ a\end{bmatrix}.

Therefore,

\displaystyle \begin{aligned}\mathbf A^{-1} &= \begin{bmatrix} \mathbf A^{-1} \mathbf e_1 & \mathbf A^{-1} \mathbf e_2 \end{bmatrix} = \frac{1}{\det(\mathbf A)} \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}.\end{aligned}

Corollary 1. For any \displaystyle a + bi = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \in \mathbb C_{\mathbb K} such that a^2 + b^2 \neq 0,

\displaystyle (a+bi)^{-1} = \frac{1}{a^2 + b^2} (a - bi).

We have listed many common computations involving 2 \times 2 matrices. Yet, a simple yet challenging problem arises: how do we compute the determinant of a 3 \times 3 matrix? More crucially, what is the determinant of such a matrix?

We are going to proceed like before. Consider a 3 \times 3 matrix

\mathbf A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}.

Our line of attack is to use the 2 \times 2 determinant to make sense of the 3 \times 3 determinant. If a_1 = b_1 = c_1 = 0, then \mathbf A is not injective and thus not invertible. We would expect \det(\mathbf A) = 0 in this setting. Now suppose a_1 \neq 0 without loss of generality. Apply elementary operations to get rid of the leading terms in the second two rows:

\begin{aligned} \mathbf A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} &\longrightarrow \begin{bmatrix} a_1 & a_2 & a_3 \\ a_1b_1 & a_1b_2 & a_1b_3 \\ a_1c_1 & a_1c_2 & a_1c_3 \end{bmatrix}\\ &\longrightarrow \begin{bmatrix} a_1 & a_2 & a_3 \\ 0 & a_1b_2 - a_2b_1 & a_1b_3 - a_3b_1 \\ 0 & a_1c_2 - a_2 c_1 & a_1c_3 - a_3 c_1 \end{bmatrix}. \end{aligned}

Since a_1 \neq 0, \mathbf A will be injective if and only if

\mathbf M := \begin{bmatrix} a_1b_2 - a_2b_1 & a_1b_3 - a_3b_1 \\ a_1c_2 - a_2 c_1 & a_1c_3 - a_3 c_1 \end{bmatrix}.

is. Yet, the entries of this matrix look very, very peculiar! Indeed, each entry is itself the determinant of some matrix. For instance,

a_1 b_2 - a_2 b_1 = \left| \begin{matrix} a_1 & a_2 \\ b_1 & b_2\end{matrix} \right|.

Let’s bravely compute the determinant of \mathbf M:

\begin{aligned}\det(\mathbf M) &= (a_1b_2 - a_2b_1) (a_1c_3 - a_3 c_1 ) - (a_1b_3 - a_3b_1)(a_1c_2 - a_2 c_1) \\ &= a_1^2 (b_2c_3 - b_3c_2) - a_1 a_2(b_1 c_3 - b_3 c_1) + a_1a_3(b_1 c_2 - b_2c_1) \\ &= a_1 \cdot \left( a_1 \left| \begin{matrix} b_2 & b_3 \\ c_2 & c_3\end{matrix} \right| - a_2 \left| \begin{matrix} b_1 & b_3 \\ c_1 & c_3\end{matrix} \right| + a_3 \left| \begin{matrix} b_1 & b_2 \\ c_1 & c_2\end{matrix} \right| \right). \end{aligned}

Then \det(\mathbf M) \neq 0 if and only if

a_1 \left| \begin{matrix} b_2 & b_3 \\ c_2 & c_3\end{matrix} \right| - a_2 \left| \begin{matrix} b_1 & b_3 \\ c_1 & c_3\end{matrix} \right| + a_3 \left| \begin{matrix} b_1 & b_2 \\ c_1 & c_2\end{matrix} \right| \neq 0.

The expression on the left-hand side turns out to be the determinant of the 3 \times 3 matrix. This yields what is known as the cofactor expansion of the matrix. Now contrary to usual practice, we won’t be defining the determinant of a 3 \times 3 matrix formally, since we have too many questions to resolve:

  • How do we generalise the definition of the determinant to n \times n matrices?
  • Does the invertibility property still hold in the general case?
  • What can we actually compute using determinants?
  • What properties do determinants actually possess?
  • What is a determinant?

The next few posts aim to resolve these questions. Once we actually can define the determinant of a general square matrix \mathbf A : \mathbb K^n \to \mathbb K^n, then we can practically discuss one of its most useful applications—eigenvectors and eigenvalues.

—Joel Kindiak, 4 Mar 25, 1815H

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