Determinants Test - 30

Given, $$A^{2}+kI = 0$$
$$A^{2} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} { a }^{ 2 }+bc & b(a+d) \\ c(a+d) & d^{ 2 }+bc \end{bmatrix}$$
If we take a $$2 \times 2$$ identical matrix $$I$$, then $$ \displaystyle kI = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$$
From equation $$A^{2}+kI=0$$, we get $$A^{2}= -kI \Rightarrow A^2= \begin{bmatrix} -k & 0 \\ 0 & -k \end{bmatrix}$$
$$a^{2}+bc = -k....(1)$$,
$$b(a+d) = 0....(2)$$
$$c(a+d) = 0....(3)$$,
$$d^{2}+bc = -k....(4)$$,

$$\begin{vmatrix}
a &b  &c \\ 
 b&c  &a \\ 
 c&a  &           b        
\end{vmatrix}$$
$$=0$$ 
$$R_{1}\rightarrow  R_{1} +R_{2}+R_{3}$$
$$\begin{vmatrix}
(a+b+c) &(a+b+c)  &(a+b+c) \\
 b&c  &a \\
 c&a  &b
\end{vmatrix}$$
$$=0$$
$$\begin{vmatrix}
 1&  1& 1\\
 b&  c&a \\   
 c&a  &b
\end{vmatrix}$$
$$=0$$
$$a+b+c =0\ ,   b, c,a > 0$$
$$\therefore a=b=c = 0$$

Given $$f(x)= \tan x,$$So,
Let, $$A=\begin{vmatrix}
f(A) & f(\pi/4) & f(\pi/4)\\
f(\pi/4) & f(B) & f(\pi/4)\\
f(\pi/4) & f(\pi/4) & f(c)
\end{vmatrix}$$
$$A=\begin{vmatrix}
\tan A & 1 & 1\\
1 & \tan B & 1\\
1 & 1 & \tan c
\end{vmatrix}$$
So, $$A=\tan A(\tan B \tan c-1)-1(\tan c-1)+1(1-\tan B)$$
$$=\tan A \tan B \tan C +2 -\tan A-\tan C -\tan B$$
By property of triangle  $$\Delta ABC$$
$$\angle (A+B+C) =180 ^{\circ}$$
So, $$\tan (A+B+C)=0$$
So, $$\tan A \tan B \tan C=\tan A + \tan C +\tan B$$
$$\therefore A=2+0=2$$

Give, $$ax + by + c = 0$$

$$2x-ay+1 =0,\ 3x-by+1 =0,\ 4x-cy+1 =0$$

Let $$ A=\begin{vmatrix}
x & y & z\\
p & q & r\\
yz & zx & xy
\end{vmatrix}$$
So, $$A= x(qxy-rzx)-y(pxy-ry z)+z(pxz-qyz)$$
$$=qx^2y-rx^2z-Pxy^2+ry^2z+pxz^2-qyz^2$$
$$=x^{p}(z^2-y^2)+qy(x^2-z^2)+rz(-x^2+y^2)$$

We know that concurrent lines have same intersection points.Then

$$\therefore 3x+2y-5=0$$ --(1)

$$2x-5y+3=0$$ ---(2)

$$5x+by+c=0$$ ---(3)

From 1 & 2,

$$6x+4y-10=0$$  

$$6x-15y+9=0$$ 

 $$\Rightarrow 19y-19=0$$

$$y=1$$ & $$x=1$$

So, $$substituting\ \  x=1$$ & $$y=1$$, in equation (3),

$$5x+by+c=0$$

$$5+b+c=0$$

$$\therefore b+c=-5$$

$$  {\textbf{Step -1: Find equation}} $$

                $$  {\text{Since A,O,C are linear they will lie on same line}}{\text{.}} $$

                $$  {\text{Given points are A}}(1,2),{\text{O}}(0,0)\;{\text{and C}}(a,b) $$

                $$  {\text{Therefore,}} $$

                $$  {\text{area }}\Delta {\text{AOC = 0}} $$

                $$ \therefore \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$

                $$  {\text{Here, }}{x_1} = 1,{x_2} = 0\;{\text{and }}{x_3} = a $$

                $$  {\text{and }}{y_1} = 2,\;{y_2} = 0\;{\text{and }}{y_3} = b $$

$$  {\textbf{Step -2: Find the relation}} $$

                $$  \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$

                $$   \Rightarrow \dfrac{1}{2}\left[ {1\left( {0 - b} \right) + 0\left( {b - 2} \right) + {\text{a}}\left( {2 - 0} \right)} \right] = 0 $$

                $$   \Rightarrow  - {\text{b + 2a}} = 0 $$

                $$   \Rightarrow {\text{2a = b}} $$

$$  {\textbf{Hence, the correct answer is option C}}{\text{.}} $$

 

 

For the points $$AB =(x, y), BC =(1,2)$$ and $$ CD=(7,0)$$ to be collinear they must have area $$=0$$.

Thus, area of a triangle ABC = $$ \frac { 1 }{ 2 } \left( x_{ 1 }\left( y_{ 2 }-y_{ 3 }

\right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 }

\right)  \right)= 0 $$

$$=\dfrac{1}{2}(2x\times 2+6y-14)=0$$

$$=(2x+6y-14)=0$$

Thus, we have the relation $$x+3y=7$$ for the points to be collinear.