We know the area of a triangle given by points $\left(x_1, y_1\right)$,$\left(x_2, y_2\right)$ $\left(x_3, y_3\right)$

in $\frac{1}{2}\left(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)\right.$ $+x_3\left(y_1-y_2\right)$ ?

Consider the following

matrix : $\quad A=\left(\begin{array}{lll}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3\end{array}\right)$

$\begin{aligned}|A|=1 & \left(x_2 y_3-x_3 y_2\right) -1\left(x_1 y_3-x_3 y_1\right) +1 \left(x_1 y_2-x_2 y_1\right)\end{aligned}$

$= x_2 y_3-x_3 y_2 - x_1 y_3 + x_3 y_1 + x_1 y_2 - x_2 y_1$

$= x_1 (y_2-y_3)+ x_2 (y_3-y_1) + x_3 (y_1-y_3)$

$\therefore$ If we compare with the formula for area of a triangle

We get

$\Delta = \frac {1}{2}$ $\left|\begin{array}{lll}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3\end{array}\right|$

$= \frac {1}{2}$ $\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|$

If determinant comes out to be-ve, we take the absolute value for area

What happens if the determinant in 0

Area of a triangle = 0 if the points are collinear

$\therefore$ To test collinearily we may use determinant in the above way

Show that $\begin{aligned} \quad & A=(a, b+c) \\ & B=(b, c+a) \\ & C=(c, a+b)\end{aligned} \quad$ are Collinear

$\therefore$ We compare with the determinant?

$\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$

$c_3=c_3+c_2$

$=$ $\left|\begin{array}{lll}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{array}\right|$

$= (a+b+c)$ $\left(\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right)$

$= 0$

If points

(a, 0) (0, b) & (1, 1)

are collinear then show that a+b = ab

since there are collinear

$\left(\begin{array}{lll}1 & a & 0 \\ 1 & 0 & b \\ 1 & 1 & 1\end{array}\right)$ $= 0$

By $R_1 \leftarrow R_1 - R_2$

We get the determinant in

$\left|\begin{array}{lll}0 & R & -b \\ 1 & 0 & b \\ 1 & 1 & 1\end{array}\right|$

Now $R_3 \leftarrow R_3 - R_2$

We have $\left|\begin{array}{lll}0 & R & -b \\ 1 & 0 & b \\ 0 & 1 & 1-b\end{array}\right|$

$|A| = (-1)^{1+2}. 1. (a(1-b)-1.(-b))$

$= -1 (a-ab+b)$

So equating with 0 We have $a-ab+b= 0 or a+b=ab$

Use deterninant method to find equation of the line joining points

$A = (-2, 4), B= (2,-6)$

We know from our knowladge of coordinate geometry:

$\frac{y-4}{4+6}=\frac{x+2}{-2-2}$ or $\frac{y-y}{10}=\frac{x+2}{-4}$

or $10x+20 =-4y$+16

or $10x+4y+4 = 0$

Let x, y be any point on the line joining $(-2, 4) \& (2, -6)$

$\therefore$ $\left|\begin{array}{lll}1 & -2 & 4 \\ 1 & 2 & -6 \\ 1 & x & y\end{array}\right| = 0$

or $\left|\begin{array}{lll}0 & -4 & 10 \\ 1 & 2 & -6 \\ 1 & x & y\end{array}\right|=0 R_1 \leftarrow R_1 - R_2$

or $\left|\begin{array}{lll}0 & -4 & 10 \\ 0 & 2-x & -6-y \\ 1 & x & y\end{array}\right|=0 R_2 \leftarrow R_2 - R_3$

or $(-4)(-6-y) - 10(2-x) = 0$

or $24 + 4y - 20 + 10x = 0$

or $10x + 4y + 4 = 0$

$(-2,4) \& (2,-6)$

If the area of the triangle $A(-2,4), \quad B(2,-6)$ & $C(5, k)$ in 35 units what is the value & $k$ ?

i.e $\frac{1}{2}\left|\begin{array}{ccc}1 & -2 & 4 \\ 1 & 2 & -6 \\ 1 & 5 & k\end{array}\right|=35 \Rightarrow\left|\begin{array}{rrr}1 & -2 & 4 \\ 1 & 2 & -6 \\ 1 & 5 & k\end{array}\right|=70$

or $\left|\begin{array}{ccc}0 & -4 & 10 \\ 1 & 2 & -6 \\ 1 & 5 & k\end{array}\right|=\begin{aligned} & 70, \ R_1 \rightarrow R_1-R_2\end{aligned}$

or $\left|\begin{array}{ccc}0 & -4 & 10 \\ 1 & 2 & -6 \\ 0 & 3 & k+6\end{array}\right|=\begin{aligned} & 70, \ R_3 \rightarrow R_3-R_2\end{aligned}$

or $(-1) (-4(k+6)-3.10) = 70$

or $(-1) (-4k-24-30) = 70, or \quad 4k+24+30 = 70, or \quad 4k = 16, or \quad k = 4$

We have expresed the determinant of a matrix A as $\sum_{j=1}^n a_{1 j} \cdot(-1)^{1+i} M_{i j}$

Where $M_{1j}$ in the determinant of the submatrix of A after deleting Row 1 & column J

In a similar way we can also write :

$|A| = \sum_j a_{i j}(-1)^{i+j} M_{i j}$

by expanding along row j This $M_{ij}$ in called the Minor corresponding the element $a_{ij}$

Consider the matrix

$A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$

$|A| = ad-bc$

$\therefore M_{11} = d \\ M_{12} = c \\ M_{21} = b \\ M_{22} = a$

But $A_{11}=(-1)^{1+1} M_{11} = d$

$A_{12}=(-1)^{1+2} M_{12} = -c$

$A_{21}= -b$

$A_{22}= a$

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

which of the following given |A|?

(A) $a_{21} M_{21}+a_{22} M_{22}+a_{23} M_{23}$

(B) $a_{11} M_{11}-a_{12} M_{21}+a_{13} M_{31}$

(C) $a_{31} A_{31}-a_{32} A_{32}+a_{33} A_{33}$

(D) $a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33}$

Answer = (D)

Def: The adjoint of a square matrix $A$ is defined as the transpose of (( $\left.A_{i j}\right)$ ) where $A_{i j}$ in the cofacter of $G_{i j}$.

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

Then adj (A) is the 3 $\times$ 3 Matrix :

$\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]$

If A = $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

Then

$A_{11} = d$ $\quad$ $A_{12} = -c$

$A_{21} = -b$ $\quad$ $A_{22} = a$

$\therefore$ $\operatorname{adj}(A)=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$

i.e the diagonal element of A are interchange & 2 sign off- off-dingonal elements are changes

Thus for a $2 \times 2$ matrix we get the adjoint very easily

What is A .(adj (A))?

$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \cdot\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$

$=\left[\begin{array}{cc}a d-b c & -a b+a b \\ c d-c d & a d-b c\end{array}\right]$

$=\left[\begin{array}{cc}|A| & 0 \\ 0 & |A|\end{array}\right]$

i.e it is a diagonal matrix with diagonal entries $=|A|$

Let us verify the result for $3 \times 3$ matrix

Let $P$ be the product of P = A (adj(A))

i. $e \quad P=\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)\left(\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right)$

We want to verify the

$P=|A|\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$

Let us look at the

$\begin{aligned} P_{11} & =a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \ & =|A|\end{aligned}$

Similarly $P_{22}$ & $P_{33}$ can be shown to be |A|

We need to show that off diagonal elements are 0 . $p_{12}=a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}$

$=a_{11}(-1)^{2+1}\left|\begin{array}{ll}a_{12} & a_{13} \\ a_{32} & a_{33}\end{array}\right|$

$+a_{12}(-1)^{2+2}\left|\begin{array}{ll}a_{11} & a_{13} \\ a_{31} & a_{33}\end{array}\right|$

$+a_{13}(-1)^{2+3}\left|\begin{array}{ll}a_{11} & a_{12} \\ a_{31} & a_{32}\end{array}\right|$

$= -a_{11} a_{12} a_{33}+a_{11} a_{13} a_{32}$ $+{a_{12} a_{11} a_{33}}{-a_{12} a_{13} a_{31} }$ $\begin{aligned} & -a_{13} a_{11} a_{32} \ +a_{13} a_{31} a_{12} \end{aligned}$ = 0

Verfied for $P_{12}$

$\therefore$ A (adj A) = |A| $I_{3}$