Ex $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$

Then $|A|=1.4-3.2$

$=4-6=-2$

Ex $A=\left[\begin{array}{ll}a+b & c+b \\ c-b & a-b\end{array}\right]$

$\begin{aligned}|A| & =(a+b)(a-b)-(c+b)(c-b) \& =a^2-b^2-\left(c^2-b^2\right) \\ & =a^2-c^2\end{aligned}$

Ex $A=\left[\begin{array}{ll}\cos \theta & \sin \theta \ -\sin \theta & \cos \theta\end{array}\right]$

Then $|A|=\cos ^2 \theta-\sin \theta(-\sin \theta)$

$\begin{aligned} & =\cos ^2 \theta+\sin ^2 \theta \& =1\end{aligned}$

Ex: $A=\left|\begin{array}{ll}1 & 2 \ 2 & 4\end{array}\right|$

Then $|A|=1.4-2.2=4-4$ $=0$

If we are expanding along the $i^{\text {th }}$ row, then

$\begin{aligned}|A|=(-1)^{i+1} & a_{i 1}\left|M_{i 1}\right| \& +(-1)^{i+2} a_{i 2}\left|M_{i 2}\right| \ +(-1)^{i+n} a_{i n}\left|M_{i n}\right| \end{aligned}$

Let me now expand along the $3^{rd}$ column

$\begin{aligned} & A=\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 1 & 2\end{array}\right) \end{aligned}$

$\begin{aligned} & |A|=(-1)^{1+3} 3(4.1-3.5) \\ & +(-1)^{2+3}.6 \cdot(1.1-3.2) \\ & \quad+(-1)^{3+3} 2(1.5-2.4) \\ & =3(4-15)-6(1-6)+2(5-8) \\ & =-33+30-6=-9\end{aligned}$

Consider a $3 \times 3$ matrix

$A=\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right)$

Its determinant when expanded along row 1 is:

$\begin{aligned} & a_{11}(a_{22} a_{33}-a_{23} a_{32}) \\ &-a_{12}(a_{21} a_{33}-a_{23} a_{31}) \\ &+a_{13}(a_{21} a_{32}-a_{22} a_{31}) \end{aligned}$

$=a_{11} a_{22} a_{33}-a_{11} a_{23} a_{32}$

$-a_{12} a_{21} a_{33}+a_{12} a_{23} a_{31}$

$+a_{13} a_{21} a_{32}-a_{13} a_{22} a_{31}$

Expand along row 3 we have

$\begin{aligned}&|A|=a_{31}(a_{12} a_{23}-a_{13} a_{22}) -a_{32}(a_{11} a_{23}-a_{13} a_{21}) +a_{33}(a_{11} a_{22}-a_{12} a_{21}) \\ &= a_{12} a_{23} a_{31}-a_{13} a_{22} a_{31} +a_{11} a_{23} a_{32} +a_{13} a_{31} a_{32} +a_{11} a_{22} a_{33} - a_{12} a_{21} a_{33}\end{aligned}$

Expansion along the $2^{nd}$ Column: $\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right)$

$|A|$ : Expanding along $2^{nd}$ column. $\begin{aligned}&\quad -a_{12}(a_{21} a_{33}-a_{23} a_{31}) \quad +a_{22}(a_{11} a_{33}-a_{13}a_{31}) \quad -a_{32}(a_{11} a_{23}-a_{13} a_{21}) \\ &=-a_{12} a_{21} a_{33}+a_{12} a_{23} a_{31} \quad +a_{11} a_{22} a_{33}-a_{13} a_{22} a_{31} \quad -a_{11} a_{23} a_{32}+a_{13} a_{21} a_{32} \end{aligned}$

Properties 1: $|A|=|A^{T}|$

For $2 \times 2$ we can easily verify:

$\begin{aligned} & A=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=a d-b c \\ & A^{T}=\left(\begin{array}{ll} a & c \\ b & d \end{array}\right) \\ &|A^{T}|=a d-b c \end{aligned}$

Determinant of a diagonal matrix is the product of its diagonal elements

$A=\left(\begin{array}{ccc}a_{11} & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{33}\end{array}\right)$

$|A| = a_{1 1} (a_{2 2} a_{3 3} - 0) - 0() + 0()$

$= a_{1 1} a_{2 2} a{3 3}$

The determinant of a triangular matrix in the product of its diagonal elements

$A=\left[\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33}\end{array}\right]$

$|A| = a_{1 1} (a_{2 2} a_{3 3} - 0.a_{ 2 3}) - 0 () + 0 ()$

$= a_{1 1} a_{2 2} a_{3 3}$.

If we multiply each element of a row (or column) by a constant k then the determinant will also be multiplied by k

i.e; if $A=\left[\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$

$B=\left[\begin{array}{ccc}k a_{11} & k a_{12} & k a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$

Then |B| is

$k a_{1 1} (a_{2 2} a_{2 3} - a_{2 3} a_{3 2})$

$- k a_{1 2} (a_{2 1} a_{3 3} - a_{3 1} a_{2 3})$

$+ k a_{1 3} (a_{2 1} a_{3 2} - a_{2 2} a_{3 1})$

$= k (a_{1 1} (a_{2 2} a_{2 3} - a_{2 3} a_{3 2}) - a_{1 2} (a_{2 1} a_{3 3} - a_{3 1} a_{2 3}) + a_{1 3} (a_{2 1} a_{3 2} - a_{2 2} a_{3 1}))$

$= k |A|$