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Determinants Test - 62

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Determinants Test - 62
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  • Question 1
    1 / -0
    If [x] stands greatest integer $$\leq x$$ then the value of
    $$\begin{vmatrix}
    \left [ e \right ] & \left [ \pi  \right ] & \left [ \pi ^{2}-6 \right ]\\
    \left [ \pi  \right ] & \left [ \pi ^{2}-6 \right ] & \left [ e \right ]\\
    \left [ \pi ^{2}-6 \right ] & \left [ e \right ] & \left [ \pi  \right ]
    \end{vmatrix}$$ equals to=?
    Solution
    $$\left[ x \right] $$Greatest integer
    $$=\begin{vmatrix} \left[ e \right]  & \left[ \pi  \right]  & \left[ { \pi  }^{ 2 }-6 \right]  \\ \left[ \pi  \right]  & \left[ { \pi  }^{ 2 }-6 \right]  & \left[ e \right]  \\ \left[ { \pi  }^{ 2 }-6 \right]  & \left[ e \right]  & \left[ \pi  \right]  \end{vmatrix}$$

    $$=\begin{vmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 3 & 2 & 3 \end{vmatrix}$$

    $$=2(5)-3(3)+3(-3)$$

    $$=-8$$
  • Question 2
    1 / -0
    If $$D_k = \begin{vmatrix}1 & n & n\\ 2k & n^2 + n + 1 & n^2 + n\\ 2k-1 & n^2 & n^2 + n + 1\end{vmatrix} $$ and $$\displaystyle \sum_{k = 1}^n D_k = 56 $$ then n equals
    Solution
    $$D_k = \begin{vmatrix}1 & n & n\\ 2k & n^2 + n + 1 & n^2 + n\\ 2k-1 & n^2 & n^2 + n + 1\end{vmatrix} $$

    $$\sum _{ k=1 }^{ n }{ D_{ k } } =\begin{vmatrix} \sum _{ k=1 }^{ n }{ 1 }  & n & n \\ 2\sum _{ k=1 }^{ n }{ k }  & n^{ 2 }+n+1 & n^{ 2 }+n \\ 2\sum _{ k=1 }^{ n }{ k } -1 & n^{ 2 } & n^{ 2 }+n+1 \end{vmatrix}$$

    $$\Rightarrow 56=\begin{vmatrix} n & n & n \\ n(n+1) & n^{ 2 }+n+1 & n^{ 2 }+n \\ n^{ 2 }+n-1 & n^{ 2 } & n^{ 2 }+n+1 \end{vmatrix}$$

    Applying $$C_{1}\rightarrow C_{1}-C_{2}$$ and $$C_{2}\rightarrow C_{2}-C_{3}$$

    $$\Rightarrow 56=n\begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & n^{ 2 }+n \\ n-1 & -(n+1) & n^{ 2 }+n+1 \end{vmatrix}$$

    $$\Rightarrow 2n=56$$
    $$\Rightarrow n=28$$
  • Question 3
    1 / -0
    If $$\Delta =\begin{vmatrix}
    x+1 & x+2 & x+a\\
    x+2 & x+3 & x+b\\
    x+3 & x+4 & x+c
    \end{vmatrix}=0$$, then
    the family of lines $$ax+by+c=0$$ passes through
    Solution
    $$\Delta =\begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}=0$$

    Applying $${ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }$$

    $$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \\ x+2 & 1 & b-2 \\ x+3 & 1 & c-3 \end{vmatrix}=0$$

    Applying $${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$$

    $$\Delta =\begin{vmatrix} x+1 & 1 & a-1 \\ 1 & 0 & b-a-1 \\ 2 & 0 & c-a-2 \end{vmatrix}=0$$

    Expanding along $${ C }_{ 2 }$$
    $$1\left( c-a-2 \right) -2\left( b-a-1 \right) =0\\ \Rightarrow a-2b+c=0$$
    From options for point $$(1,-2)$$ lies on the line $$ax+by+c=0$$
  • Question 4
    1 / -0
    Consider the points $$P=(-\sin (\beta -\alpha ), -\cos \beta )$$, $$Q=(\cos (\beta -\alpha ), \sin \beta )$$ and $$R=(\cos (\beta -\alpha +\theta ), \sin (\beta -\theta ))$$, where $$0< \alpha , \beta < \dfrac{\pi }{4}$$ then 
    Solution
    $$\Delta=\displaystyle \begin{vmatrix} x_1 & y_1  & 1 \\ x_2  & y_2  & 1 \\ x_3 & y_3  & 1  \end{vmatrix}$$

    Put $$\beta -\alpha =\phi $$ and consider the determinant
    $$\Delta =\displaystyle \begin{vmatrix} -\sin \phi  & -\cos \beta  & 1 \\ \cos \phi  & \sin \beta  & 1 \\ \cos (\phi +\theta ) & 
    \sin (\beta -\theta )  & 1 \end{vmatrix}$$
    Using $$R_{3}\rightarrow R_{3}-\cos \theta R_{2}-\sin \theta R_{1}$$
    $$\Delta =\displaystyle \begin{vmatrix} -\sin \phi  & -\cos \beta  & 1 \\ \cos \phi  & \sin \beta  & 1 \\ 0 & 0  & 1-\cos \theta 
    -\sin \theta  \end{vmatrix}$$
    $$=(1-\cos \theta -\sin \theta )\cos (\phi +\beta )$$
    $$=(1-\cos \theta -\sin \theta )\cos (2\beta -\alpha )$$
    $$=\left [ 1-\sqrt{2}\:\sin (\theta +\dfrac{\pi}4) \right ]\cos (2\beta -\alpha )$$
    As $$0< \theta < \pi /4\Rightarrow \dfrac{\pi}4< \theta +\dfrac{\pi}4< \dfrac{\pi}2$$
    $$\Rightarrow \displaystyle \frac{1}{\sqrt{2}}< \sin (\theta +\dfrac{\pi}4)< 1$$
    $$\Rightarrow \displaystyle \cos (2\beta -\alpha )\neq 0$$
    Thus $$\Delta \neq 0$$ and the points $$P, Q, R$$ are non-collinear.
  • Question 5
    1 / -0
    If the points $$(a, 1), (1, b)$$ and $$(a -1, b -1)$$ are collinear, $$\alpha ,\beta $$ are respectively the arithmetic and geometric means of $$a$$ and $$b $$, then $$4\alpha -\beta^{2}$$ is equal to
    Solution
    Since, The given points are collinear 
    $$\displaystyle \begin{vmatrix} x_1 & y_1  & 1 \\ x_2  & y_2  & 1 \\ x_3 & y_3  & 1  \end{vmatrix}=0$$

    so,$$\displaystyle \begin{vmatrix} a  & 1  & 1 \\ 1  & b  & 1 \\a-1& b-1  & 1  \end{vmatrix}=0$$
    On expanding,we get
    $$a(b-b+1)-1(1-a+1)+1(b-1-ab+b)=0$$
    $$2(a+b)-ab=3$$  -----------1)
    Now $$\alpha=\dfrac{a+b}{2} and\   \beta=\sqrt{ab}$$  ---------2)
    From 1) and 2)
    Now $$4\alpha-{\beta}^{2}=3$$
    $$2(a+b)-ab{=}3$$
  • Question 6
    1 / -0
    Two $$n \times n$$ square matrices $$A$$ and $$B$$ are said to be similar if there exists a non-singular matrix $$P$$ such that  $$P^{-1}A\: P=B$$
    If $$A$$ and $$B$$ are two similar matrices, then

    Solution
    As $$A$$ and $$B$$ are similar matrices there exists a non-singular matrix $$P$$ such that $$A=P^{-1}\:BP$$

    $$\Rightarrow det \: (A) = det\: (P^{-1}\:BP)$$

    $$=det\: (P^{-1})\: det \: (B) \: det \: (P)$$

    $$\displaystyle =\frac{1}{det\: (P)}det \: (B) \: (det P)$$

    $$= det \: B$$

    Hence, option A.

  • Question 7
    1 / -0
    If $$\Delta =\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ and $$c_{ij}=\left ( -1 \right )^{i+j}$$ (determinant obtained by deleting ith row and jth column), then $$\begin{vmatrix} c_{11} & c_{12} & c_{13}\\ c_{21} & c_{22} & c_{23}\\ c_{31} & c_{32} & c_{33} \end{vmatrix}=\Delta ^{2}$$


    If $$\begin{vmatrix} 1 & x & x^{ 2 } \\ x & x^{ 2 } & 1 \\ x^{ 2 } & 1 & x \end{vmatrix}=7$$ and $$\Delta =\begin{vmatrix}
    x^{3}-1 & 0 & x-x^{4}\\
    0 & x-x^{4} & x^{3}-1\\
    x-x^{4} & x^{3}-1 & 0
    \end{vmatrix}$$, then
    Solution
    For $$\begin{vmatrix} 1 & x & x^{ 2 } \\ x & x^{ 2 } & 1 \\ x^{ 2 } & 1 & x \end{vmatrix}=7$$
    $$\begin{vmatrix} c_{ 11 } & c_{ 12 } & c_{ 13 } \\ c_{ 21 } & c_{ 22 } & c_{ 23 } \\ c_{ 31 } & c_{ 32 } & c_{ 33 } \end{vmatrix}=\begin{vmatrix} x^{ 3 }-1 & 0 & x-x^{ 4 } \\ 0 & x-x^{ 4 } & x^{ 3 }-1 \\ x-x^{ 4 } & x^{ 3 }-1 & 0 \end{vmatrix}$$
    $$\Delta ={ 7 }^{ 2 }=49$$
  • Question 8
    1 / -0
    The number of distinct real roots of $$\begin{vmatrix} \sin\, x&\cos\, x&\cos \,x \\ \cos\, x &\sin\, x&\cos\, x \\ \cos\, x&\cos\, x&\sin \, x\end{vmatrix}=0$$ in the interval $$-\dfrac{\pi}{4} < x \le \dfrac{\pi}{4}$$ is
    Solution
    Given, $$\begin{vmatrix} \sin\, x&\cos\, x&\cos \,x \\\cos\, x &\sin\, x&\cos\, x \\\cos\, x&\cos\, x&\sin \, x\end{vmatrix}=0$$ 

    $$\Rightarrow \sin \,x(\sin^2x-\cos^2x)-\cos\,x(\sin\,x\,\cos\,x-\cos^2x)+\cos\,x(\cos^2x-\sin\,x\,\cos\,x)=0$$

    $$\Rightarrow \sin \,x(\sin\,x-\cos\,x)(\sin\,x+\cos\,x)-\cos^2\,x(\sin\,x-\cos\,x)-\cos^2\,x(\sin\,x-\cos\,x)=0$$

    $$\Rightarrow (\sin\,x-\cos\,x)(\sin^2x+\sin\,x\,\cos\,x-2\cos^2\,x)=0$$

    $$\Rightarrow (\sin\,x-\cos\,x)(\sin^2\,x+2\,\sin\,x\,\cos\,x-\sin\,x\,\cos\,x-2\,\cos^2\,x)=0$$

    $$\Rightarrow (\sin\,x-\cos\,x)^2(\sin\,x+2\,\cos\,x)=0$$

    $$\Rightarrow \sin\,x-\cos\,x=0$$ or $$\sin\,x+2\,\cos\,x=0$$

    $$\Rightarrow \tan\, x=1$$ or $$\tan\, x=-2$$

    But $$x \in\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$$
    $$\Rightarrow$$ Only one solution i.e $$x=\dfrac{\pi}{4}$$ 
  • Question 9
    1 / -0
    If the points $$(-1,3), (2,p)$$ and $$(5,-1)$$ are collinear, the value of $$p$$ is
    Solution
    Let $$A(-1,3)$$, $$B(2,p)$$ and $$C(5,-1)$$ be the given points.
    Three points are collinear if $${ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 })=0$$
    Therefore, $${ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 })=0\\ \Rightarrow -1(p-(-1))+2(-1-3)+5(3-p)=0\\ \Rightarrow -1(p+1)+2(-4)+5(3-p)=0\\ \Rightarrow -p-1-8+15-5p=0\\ \Rightarrow -6p+6=0\\ \Rightarrow -6p=-6\\ \Rightarrow p=1$$
    Hence $$p=1$$
  • Question 10
    1 / -0
    $$(1,6), (3.-2)$$ and $$(-2,K)$$ are collinear points. What is $$K$$?
    Solution
    If three points are collinear then the area of triangle is zero.

    Therefore, if the points $$(1,6)$$, $$(3,-2)$$ and $$(-2,K)$$ are collinear, then their area must be zero that is:
    $$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 1(-2-K)+3(K-6)-2(6+2) \right| \\ \Rightarrow 0=\left| -2-K+3K-18-16 \right| \\ \Rightarrow 0=\left| 2K-36 \right| \\ \Rightarrow 2K-36=0\\ \Rightarrow 2K=36\\ \Rightarrow K=18$$

    Hence, $$K=18$$
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