$2 \times 2$ case,
$$\det[A]\ = \ \left |\begin{array}{cc} a & b \\ c & d \end{array}\right |
\ =\ ad\,-\,bc\,.$$We need to define determinants and list a few of their principal properties. You've probably met them already in high school and in calculus as well as in differential equations:
$2 \times 2$ case,
$$\det[A]\ = \ \left |\begin{array}{cc} a & b \\ c & d \end{array}\right |
\ =\ ad\,-\,bc\,.$$
$3 \times 3$ case,
$$\det [A]\ = \ \left|\begin{array}{ccc} X & Y & Z \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|
\ = \ X \left|\begin{array}{cc} a_2 & a_3 \\ b_2 & b_3 \end{array}\right|
- Y \left|\begin{array}{cc} a_1 & a_3 \\ b_1 & b_3 \end{array}\right|
+ Z \left|\begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \end{array}\right|
\qquad\qquad \qquad $$
$$\qquad \qquad \qquad = \ X(a_2b_3 - b_2a_3) - Y(a_1b_3 - b_1a_3) + Z(a_1b_2 - b_1a_2)\,,$$
the entries being labeled solely to emphasize how they get combined
and multiplied.
For example, in the $2 \times 2$ case $$\left |\begin{array}{cc} 4 & 2 \\ 3 & 1 \end{array}\right | \ =\ 4 \cdot 1\,-\,3 \cdot 2 \ = \ -2\,,$$ while $$\left|\begin{array}{ccc} 1 & 2 & -3 \\ 4 & 1 & 1 \\ 1 & 2 & -1 \end{array}\right| \ = \ \left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right| - 2 \left|\begin{array}{cc} 4 & 1 \\ 1 & -1 \end{array}\right| -3 \left|\begin{array}{cc} 4 & 1 \\ 1 & 2 \end{array}\right|\ = \ -3 -2(-5) -3(7) \ = \ -14\,. $$
This way of defining $3 \times 3$ determinants in terms of sums of signed multiples of $2 \times 2$ determinants can be continued to a general $n \times n$ matrix. The $4 \times 4$ case is typical:
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Given a $4 \times 4$ matrix $$ A \ = \ \left[\begin{array}{cc} 2 & 7 & -1 & 5 \\ -3 & 4 & 2 & 0 \\ 3 & 1 & -2 & 4 \\ 0 & 2 & 5 & 3 \end{array} \right] \ = \ \big[\, {\bf a}_1 \ \ {\bf a}_2 \ \ {\bf a}_3 \ \ {\bf a}_{4} \big] \,,$$ we define $3 \times 3$ sub-matrices $A_{1k}$ by deleting the first row and ${\bf a}_k$ column of $A$, and then define $\det[A]$ by $$\det[A]\ = \ a_{11}\det[A_{11}] \,-\, a_{12}\det[A_{12}]\, + \, a_{13}\det[A_{13}] \,-\, a_{14}\det[A_{14}] \qquad \qquad \qquad \qquad $$ $$ = \ 2\,\text{det}\left[\begin{array}{cc}4 & 2 & 0 \\ 1 & -2 & 4 \\ 2 & 5 & 3 \end{array}\right] - 7\, \text{det}\left[\begin{array}{cc}-3 & 2 & 0 \\ 3 & -2 & 4 \\ 0 & 5 & 3 \end{array}\right] - \text{det}\left[\begin{array}{cc}-3 & 4 & 0 \\ 3 & 1 & 4 \\ 0 & 2 & 3 \end{array}\right] -5\, \text{det}\left[\begin{array}{cc}-3 & 4 & 2 \\ 3 & 1 & -2 \\ 0 & 2 & 5 \end{array}\right]\,. $$ |
Lengthy calculations show that $\det[A] \,= \,-212 $.
Problem: evaluate the determinant of the $4 \times 4$ matrix $$B \ = \ \left[\begin{array}{cc} 1 & 0 & 0 & 2 \\ 0 & 3 & 4 & 0 \\ 0 & a & b & 0 \\ c & 0 & 0 & d \end{array}\right].$$
Proceeding recursively, the determinant of an $n \times n$ matrix is defined using the determinant of $(n-1) \times (n-1)$ matrices:
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Definition: the Determinant of an $n \times n$ matrix $$A \ = \ \left[\begin{array}{cc} a_{11} & a_{12} & \cdots & a_{1,\, n} \\ a_{21} & a_{22} & \cdots & a_{2,\, n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,\, 1} & a_{n,\, 2} & \cdots & a_{n, \, n} \end{array} \right] \ = \ \big[\, {\bf a}_1 \ \ {\bf a}_2 \ \ \cdots \ \ {\bf a}_{n} \big] \,,$$ is the alternating sum $$\det[A] \ = \ a_{11}\det[A_{11}] - a_{12} \det[A_{12}] + \ldots + (-1)^{n-1}a_{1,\, n}\det[A_{1,\, n}]$$ where the sub-matrix $A_{1k}$ is obtained from $A$ by deleting the first row and ${\bf a}_k$ column of $A$. |
Now to some properties:
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Properties of Determinants 1. when $A,\ B$ are arbitrary $n \times n$ matrices, This can be checked in the $2 \times 2$ case using direct calculation: since $$A\,B \,=\, \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]\,\left[\begin{array}{cc} p & q \\ r & s \end{array}\right] \,=\, \left[\begin{array}{cc} ap+br & aq+bs \\ cp+dr & cq+ds \end{array}\right],$$ $$\det[A\,B] \ = \ adps - adrq - bcps + bcrq \qquad $$ $$\qquad = \ (ad-bc)(ps-rq) \,=\, \det[A]\,\det[B]\,. $$ |
Properties of Determinants 2. when $A$ is an invertible $n \times n$ matrix, When $A$ is invertible, then its inverse is an $n \times n$ matrix $A^{-1}$ such that $AA^{-1} \,=\, I_n$. But then $$\det[A]\,\det[A^{-1}] \ = \ \det[A A^{-1}] \ = \ \det[I_n] \ = \ 1\,. $$ |
Here are two results that can be established with these properties. They will become important shortly.
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Fundamental Theorem 1: an $n \times n$ matrix $A$ is invertible if and only if $\det\, A \,\ne \, 0$. |
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Fundamental Theorem 2: if $B$ is an $n \times n$ matrix, then the homogeneous equation $B\, {\bf x} \,=\, {\bf 0}$ has non-trivial solutions if and only if $\det\,B \,= \, 0$. |