Matrix determinants in general

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 The determinant of a general \(n\times n\) matrix is quite complicated. We give a mechanical, recursive definition first and then think about its meaning later.



Definition (Determinant)

Let \(A = (a_{ij}) \in M_n\). The determinant \(|A|\) of \(A\) is defined recursively in the following manner.

  • If \(n = 1\), then \(|A| = a_{11}\).
  • If \(n > 1\), then let \(A_{ij}\) denote the \((n-1)\times(n-1)\) matrix obtained by removing the \(i\)-th row and the \(j\)-th column from \(A\), and \[|A| = \sum_{j=1}^{n}(-1)^{i+j}a_{ij}|A_{ij}|\] where \(i\) is any arbitrary index from 1 to \(n\). (Instead of a row, you may use an arbitrary column to obtain the same result.)
According to this definition, if we want to compute \(|A|\) of an \(n\times n\) matrix, we need to compute the determinants \(|A_{ij}|\) of (many) \((n-1)\times(n-1)\) matrices, which requires computing determinants of \((n-2)\times(n-2)\) matrices, and so on, until we reach the determinants of \(1\times 1\) matrices which are trivial to compute (actually we already know how to compute the determinant of a \(2\times 2\) matrix so we may stop at \(n=2\)).

Example. First, let's see the previous definition of the determinant of a \(2\times 2\) matrix is consistent with the above definition.
Let \(A = \begin{pmatrix}a & b\\c & d\end{pmatrix}\). Then,
\[|A| = (-1)^{1+1}a|(d)| + (-1)^{1+2}b|(c)| = ad - bc.\] Thus, the two definitions are indeed consistent. □

Example. 
Let
\[ A = \begin{pmatrix} 1 & 0 & 2\\ -1 & 1 & 1\\ 0 & 1 & 0 \end{pmatrix} = (a_{ij}). \]
To compute its determinants, let's pick the first row as ``\(i\)''. So
\[ \begin{eqnarray*} |A| &=& (-1)^{1+1}a_{1,1} \begin{vmatrix} a_{2,2} & a_{2,3}\\ a_{3,2} & a_{3,3} \end{vmatrix} + (-1)^{1+2}a_{1,2} \begin{vmatrix} a_{2,1} & a_{2,3}\\ a_{3,1} & a_{3,3} \end{vmatrix} + (-1)^{1+3}a_{1,3} \begin{vmatrix} a_{2,1} & a_{2,2}\\ a_{3,1} & a_{3,2} \end{vmatrix}\\ &=& (-1)^{1+1}\cdot 1\cdot \begin{vmatrix} 1 & 1\\ 1 & 0 \end{vmatrix} + (-1)^{1+2}\cdot 0\cdot \begin{vmatrix} -1 & 1\\ 0 & 0 \end{vmatrix} + (-1)^{1+3}\cdot 2\cdot \begin{vmatrix} -1 & 1\\ 0 & 1 \end{vmatrix}\\ &=& -1 + 0 + (-2) = -3. \end{eqnarray*} \]
Since we may pick any arbitrary row (\(i\)), we may equally use the second row to compute
\[ \begin{eqnarray*} |A| &=& (-1)^{2+1}(-1) \begin{vmatrix} 0 & 2\\ 1 & 0 \end{vmatrix} + (-1)^{2+2}(1) \begin{vmatrix} 1 & 2\\ 0 & 0 \end{vmatrix} + (-1)^{2+3}(1) \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix}\\ &=& -2 + 0 - 1 = -3 \end{eqnarray*} \]
to obtain the same result. □

Example. 
Let \(X = (x_{ij}) \in M_3\). Prove the following.
\[ |X| = x_{11}x_{22}x_{33} - x_{11}x_{23}x_{32} + x_{12}x_{23}x_{31} - x_{12}x_{21}x_{33} +x_{13}x_{21}x_{32} - x_{13}x_{22}x_{31}. \]
Do you see any pattern? There are \(3!=6\) terms. Each term is a product of three elements. Each of 1, 2, 3 appears only once in the first index of each factor; the same for the second index. Half (3) of the terms are multiplied by -1. What are these patterns? They are permutations! As fascinating as it is, we don't delve into the details here. See a textbook on linear algebra. □





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