Determinants Test - 37

$$\displaystyle A= \begin{bmatrix}0 &i-\sin x  &i- \cos x\\sin x -i &0  &\sin x-i \\cos x-i &- \sin x+ i  &0 \end{bmatrix}$$

$$\displaystyle \therefore \left |  A\right | = \begin{vmatrix} 0 & i-\sin x & i-\cos x\\\sin x -i  & 0  & \sin x-i \\\cos x-i  & i-\sin x & 0\end{vmatrix} $$

$$|kA|=k^{n}|A|$$ where n is order of matrix
Let $$\Delta =|A|$$
So, $${\Delta}'=|5A|=5^{3}|A|$$
$${\Delta}'=125{\Delta}$$

$${\textbf{Step1: Proving the three points are collinear  }}$$

              $${\text{Given points are A(a,b + c),B(b,c + a),C(c,a + b)}}$$

              $${\text{ Lets the points of the triangle be }}\mathbf{{({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})}}$$

               $${\text{[}}{{\text{x}}_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})] = 0$$

              $$\text{Comparing these with the given points}$$

              $$[a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)] $$

              $$=ac - ab + ab - bc + bc - ac$$

              $$ = 0$$

$${\textbf{Hence,The given points are collinear.}}$$

$$\displaystyle A=\left [ \begin{matrix}1 &-1  &1 \\ 2 &1  &-3 \\ 1 &1  &1 \end{matrix} \right ]$$
 and $$\displaystyle B=\left [ \begin{matrix}4 &2  &2 \\ -5 &0  &\alpha  \\ 1 &-2  &3 \end{matrix} \right ]$$

Also given, $$adj A=B$$

Now,$$adj A=C^{T}=\left[ \begin{matrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] ^{T}$$

$$\Rightarrow adj A=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] $$

$$\Rightarrow \left[ \begin{matrix} 4 & 2 & 2 \\  -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] =\left [ \begin{matrix}4 &2  &2 \\ -5 &0  &\alpha  \\ 1 &-2  &3 \end{matrix} \right ]$$

On comparing ,we get $$\alpha=5$$

If $$3$$ points $$P\left( { x }_{ 1 },{ y }_{ 1 } \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$ R\left( { x }_{ 3 },{ y }_{ 3 } \right) $$ are collinear, then slope $$ { m }_{ PQ }=$$ slope $$ { m }_{ QR } =$$ slope $${ m }_{ PR }$$. 

When three points A, B and C are collinear, slope of line

joining any two points (say AB) $$ = $$ slope of line joining any other two

points (say BC).

Slope of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 }

\right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ $$ = $$ $$\dfrac { { y

}_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$

Slope of the line passing through $$( \dfrac {2}{5}, \dfrac {1}{3})$$ and $$( \dfrac {1}{2}, k) 

=$$ Slope of line passing through $$(\dfrac {1}{2}, k)$$ and $$( \dfrac {4}{5},0) $$

 $$ \dfrac { k- \dfrac{1}{3} }{ \dfrac {1}{2} - \dfrac {2}{5} } = \dfrac { 0 - k}{ \dfrac {4}{5} - \dfrac{1}{2} } $$

On solving, $$ k =  \dfrac {1}{4} $$

The given points are $$ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 7,8 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -5,2 \right) $$ and $$ C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 3,6 \right) $$. 

$$x+iy=\begin{vmatrix} 1 & i & -\omega  \\ -i & 1 & \omega ^{ 2 } \\ \omega  & -\omega ^{ 2 } & 1 \end{vmatrix}\\$$

$$\displaystyle \Delta =\begin{vmatrix}
0 &b-a  &c-a \\
a-b &0  &c-b \\
a-c &b-c  &0
\end{vmatrix}$$

Area of a triangle with vertices $$({

x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y }_{ 2 })$$  and $$({

x }_{ 3 },{ y }_{ 3 })$$  is $$ \left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 }  \right| $$

Since the given points are collinear, they do not form a triangle, which means area of the triangle is zero.

Hence, substituting the points $$({

x }_{ 1 },{ y }_{ 1 }) = (x,0) $$ ; $$({ x }_{ 2 },{ y }_{ 2 }) = (0,y) $$  and $$({

x }_{ 3 },{ y }_{ 3 }) = (1,1)$$ in the area formula, we get

$$ \left| \dfrac { { x }({ y }-1)+0(1-0)+1(0-{ y }) }{ 2 }  \right|  = 0 $$
$$ => xy - x -y = 0 $$
$$ => x + y = xy $$
Hence, option 'B' is correct.