Determinants Test - 61

$$p_{1}x+q_{1}y=1,\ p_{2}x+q_{2}y =1\ p_{3}x+q_{3}y=1$$
Given lines are concurrent
$$\Rightarrow \begin{vmatrix} p_{1} & q_{1} & 1 \\ p_{2} & q_{2}& 1 \\ p_{3} & q_{3} & 1 \end{vmatrix}=0$$

$$\Rightarrow p_{1}(q_{2}-q_{3})-q_{1}(p_{2}-p_{3})+(p_{2}q_{3}-p_{3}q_{2})=0$$

$$\Rightarrow (p_{1}q_{2}-p_{2}q_{1})+(p_{2}q_{3}-p_{3}q_{2})+(p_{3}q_{1}-p_{1}q_{3})=0$$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates $$(p_1, q_1),\; (p_2, q_2),\; (p_3,q_3)$$
Since it is equal to zero, $$(p_{1},q_{1}),(p_{2},q_{2}),(p_{3},q_{3})$$ are collinear.

Given $$x+ay+a=0$$   ...(1)

$$bx+y+b=0$$   ...(2)

$$cx+cy+1=0$$    ...(3)

These lines are concurrent means they have only one intersection point

From 1 & 2,

$$(ab-1)y+ab-b=0$$

$$y=\dfrac{b-ab}{ab-1}$$ & $$x=-a\left(\dfrac{b-1}{ab-1}\right)$$

From 1 & 3,

$$(ac-c)y+ac-1=0$$

$$y=\dfrac{1-ac}{ac-c}$$ & $$x=-a\left(\dfrac{1-c}{ac-c}\right)$$

So, 

$$\dfrac{b-ab}{ab-1}=\dfrac{1-ac}{ac-c}$$

$$\Rightarrow abc-cb+abc-a^2bc=ab-a^2bc-1+ac$$

$$\Rightarrow 1+2abc=ab+ac+bc$$    ...(4)

Now,

$$\dfrac{a}{a-1}+\dfrac{b}{b-1}+\dfrac{c}{c-1}=\dfrac{ab-a+ab-b}{ab-a-b+1}+\dfrac{c}{(c-1)}$$

$$=\dfrac{2abc-ac-bc-2ab+a+b+abc-ac-bc+c}{abc-ac-bc+c-ab+a+b-1}$$

$$=\dfrac{3abc-2(ac+bc+ab)+a+b+c}{abc-(ac+bc+ab)+a+b+c-1}$$

$$=\dfrac{abc-(ac+bc+ab)+(a+b+c)-1}{abc-(ac+bc+ab)+(a+b+c)-1}=1$$ [From (4)]

Hence, option C.

Let $$ x_1=a , x_2=ar $$ and $$x_3=ar^2;  y_1=b, y_2=br$$ and $$y_3=br^2$$
Now, $$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{br-b}{ar-a}=\dfrac{b}{a}$$
Similarly, $$\dfrac{y_3-y_2}{x_3-x_2}=\dfrac{br^2-br}{ar^2-ar}=\dfrac{b}{a}$$
Thus, the points are collinear as they lie on same line. 

Given $$A= \begin{bmatrix} 1& 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix}$$ satisfies $$A^n = A^{n - 2} + A^2 - I$$ for $$n \geq 3$$
$$\Rightarrow A^4=A^2+A^2-I=2A^2-I$$
$$\Rightarrow A^6=A^4+A^2-I=3A^2-2I$$
similarly
$$A^{50}=25A^2-24I$$
$$A^2=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$
$$\therefore A^{50}=\begin{bmatrix} 25 & 0 & 0 \\ 25 & 25 & 0 \\ 25 & 0 & 25 \end{bmatrix}-\begin{bmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$
$$\therefore |A^{50}|=1$$
Hence, option B.

Given $$A= \begin{bmatrix} 1& 0 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix}$$ satisfies $$A^n = A^{n - 2} + A^2 - I$$ for $$n \geq 3$$
$$\Rightarrow A^4=A^2+A^2-I=2A^2-I$$
$$\Rightarrow A^6=A^4+A^2-I=3A^2-2I$$
similarly
$$A^{50}=25A^2-24I$$
$$A^2=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$
$$\therefore A^{50}=\begin{bmatrix} 25 & 0 & 0 \\ 25 & 25 & 0 \\ 25 & 0 & 25 \end{bmatrix}-\begin{bmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$
$$\Rightarrow A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$
Given $$\displaystyle \bigcup_{3 \times 3}$$ with its column as $$\cup_1, \cup_2, \cup_3$$ such that $$A^{50} \cup_1 = \begin{bmatrix}1\\25 \\ 25\end{bmatrix}, A^{50} \cup_2 = \begin{bmatrix}0\\ 1 \\ 0\end{bmatrix}, A^{50} \cup_3 = \begin{bmatrix}0\\ 0 \\ 1\end{bmatrix}$$
let $$B=A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$
$$B^{-1}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}$$
$$\cup_1=B^{-1}\begin{bmatrix}1\\25 \\ 25\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}\begin{bmatrix}1\\25 \\ 25\end{bmatrix}=\begin{bmatrix}1\\0 \\ 0\end{bmatrix}$$
$$\cup_2=B^{-1}\begin{bmatrix}0\\1 \\ 0\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}\begin{bmatrix}0\\1 \\ 0\end{bmatrix}=\begin{bmatrix}0\\1 \\ 0\end{bmatrix}$$
$$\cup_3=B^{-1}\begin{bmatrix}0\\0 \\ 1\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -25 & 1 & 0 \\ -25 & 0 & 1 \end{bmatrix}\begin{bmatrix}0\\0 \\ 1\end{bmatrix}=\begin{bmatrix}0\\0 \\ 1\end{bmatrix}$$
$$\therefore |\cup|=\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}=1$$
Hence, option B.

$$\left |\left (A-A^{T}  \right )^{105}  \right |=\left |-\left (A^{T}-A  \right )  \right |^{105}$$
  
           $$=-\left |A^{T}-A  \right |^{105}$$

            $$=-\left |\left (A^{T}-A  \right )^{T}  \right |^{105}$$    ($$\because $$ determinant of matrix and its transpose is equal)

            $$=-\left |A-A^{T}  \right |^{105}$$

$$\Rightarrow \left |A-A^{T}  \right |^{105}=-\left |A-A^{T}  \right |^{105}=0    (\because $$ determinant of skew-symmetric matrix of odd order is 0.) 
  
$$\left |\left (A-A^{T}  \right )^{105}  \right |=0$$
Hence, option 'D' is correct.             

Given, $$ \begin{vmatrix} 1+x  & x & x^{2}\\ x & 1+x & x^{2} \\ x^{2}  &  x & 1+x \end{vmatrix} =   ax^{5} + bx^{4} + cx^{3} + dx^{2} + \lambda x + \mu $$
Put $$x=0$$
$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}=\mu $$
$$\Rightarrow \mu=1$$
Differentiating both sides, 
$$\begin{vmatrix} 1 & 1 & 2x \\ x & 1+x & x^{ 2 } \\ x^{ 2 } & x & 1+x \end{vmatrix}+\begin{vmatrix} 1+x & x & x^{ 2 } \\ 1 & 1 & 2x \\ x^{ 2 } & x & 1+x \end{vmatrix}+\begin{vmatrix} 1+x & x & x^{ 2 } \\ x & 1+x & x^{ 2 } \\ 2x & 1 & 1 \end{vmatrix}=5ax^{ 4 }+4bx^{ 3 }+3cx^{ 2 }+2dx+\lambda $$
Put $$x=0$$
$$\lambda =\begin{vmatrix} 1&1&0 \\ 0&1&0 \\ 0&0&1 \end{vmatrix}+\begin{vmatrix} 1&0&0 \\ 1&1&0 \\ 0&0&1 \end{vmatrix}+\begin{vmatrix} 1&0&0 \\ 0&1&0 \\ 0&1&1 \end{vmatrix}$$
$$=1+1+1=3$$

Given points $$\displaystyle(-2,0),\left(-1,\dfrac1{\sqrt{3}}\right)$$ and $$\displaystyle(\cos\theta,\sin \theta)$$
For collinearity,

Given, $$\begin{vmatrix} x& 2 & x\\ x^2 & x & 6\\ x & x & 6\end{vmatrix}  = \alpha x^4 + \beta x^3 + \gamma x^2 + \delta x + \lambda$$

Put $$x=0$$
$$\begin{vmatrix} 0 & 2 & 0 \\ 0 & 0 & 6 \\ 0 & 0 & 6 \end{vmatrix}=\lambda $$
$$\Rightarrow \lambda=0$$

So, $$\begin{vmatrix} x& 2 & x\\ x^2 & x & 6\\ x & x & 6\end{vmatrix}  = \alpha x^4 + \beta x^3 + \gamma x^2 + \delta x$$
Differentiating w.r.t x,
$$\begin{vmatrix} 1 & 0 & 1 \\ x^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}+\begin{vmatrix} x & 2 & x \\ 2x & 1 & 0 \\ x & x & 6 \end{vmatrix}+\begin{vmatrix} x & 2 & x \\ x^{ 2 } & x & 6 \\ 1 & 1 & 0 \end{vmatrix}=4\alpha x^{ 3 }+3\beta x^{ 2 }+2\gamma x+\delta $$
Put $$x=0$$
$$\Rightarrow \begin{vmatrix} 1 & 0 & 1 \\ 0 & 0 & 6 \\ 0 & 0 & 6 \end{vmatrix}+\begin{vmatrix} 0 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 6 \end{vmatrix}+\begin{vmatrix} 0 & 2 & 0 \\ 0 & 0 & 6 \\ 1 & 1 & 0 \end{vmatrix}=\delta $$

$$\Rightarrow \delta=-12$$