Determinants Test - 27

$$\begin{vmatrix}
1 & 4 & 20\\
1 & -2 & 5\\
1 & 2x & 5x^2
\end{vmatrix}=0$$
$$r_1 \rightarrow r_1-r_2$$ and $$r_3\rightarrow r_3-r_2$$
$$\begin{vmatrix}
0 & 6 & 15\\
1 &  -2 & 5\\
0 & 2x+2 & 5x^2-5
\end{vmatrix}$$
$$=-(6(5x^2-5)-15(2x+2))$$
$$=-(30x^2-30-30x-30)$$
$$=30(-x^2+x+2)$$
So, $$x^2-x-2=0$$
$$x=-1,2$$

$$\begin{vmatrix} 1 & 1 & 1 \\ 1+sinA & 1+sinB & 1+sinC \\ sinA+sin^{ 2 }A & sinB+sin^{ 2 }B & sinC+sin^{ 2 }C \end{vmatrix}$$=0

$$\Rightarrow \sin { B } \sin ^{ 2 }{ C } -\sin { C } \sin ^{ 2 }{ B+ } \sin { C } \sin ^{ 2 }{ A } -\sin { A } \sin ^{ 2 }{ C } +\sin { A } \sin ^{ 2 }{ B } -\sin { B } \sin ^{ 2 }{ A } =0$$

$$\Rightarrow (\sin A-\sin B)(\sin B-\sin C)(\sin C-\sin A)=0$$

$$\Rightarrow  (\sin A-\sin B)=0$$ or $$ (\sin B-\sin C)=0$$ or $$(\sin C-\sin A)=0$$
$$\Rightarrow A=B$$ or $$B=C$$ or $$C=A$$ 

$$a=1+2+4+---$$ upto $${ n }$$ terms
$$\Rightarrow a={ 2 }^{ n }-1$$

$$b=1+3+9+---$$ upto $$ { n }$$ terms
$$\Rightarrow\displaystyle b=\dfrac{{3}^{n}-1}{2}$$

$$c=1+5+25+----$$ upto $$ { n } $$ terms
$$\Rightarrow\displaystyle c=b=\dfrac { { 5 }^{ n }-1 }{ 4 } $$

$$\Delta =\begin{vmatrix} a & 2b & 4c \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}$$

$$=\begin{vmatrix} 2^{ n }-1 & 3^{ n }-1 & 5^{ n }-1 \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}$$

$$=\begin{vmatrix} 2^{ n } & 3^{ n } & 5^{ n } \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}+\begin{vmatrix} -1 & -1 & -1 \\ 2 & 2 & 2 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}$$

$$=0-2\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2^{ n } & 3^{ n } & 5^{ n } \end{vmatrix}=0$$

$${\textbf{Step -1: Given determinant is,}}$$

                    $${\text{Let,}}$$  $$A = \left| {\begin{array}{*{20}{c}}  a&{2b}&{2c} \\   3&b&c \\  4&a&b\end{array}} \right|$$

$${\textbf{Step -2: Expand the determinant.}}$$

                  $$A = a\left( {{b^2} - ac} \right) - 2b\left( {3b - 4c} \right) + 2c\left( {3a - 4b} \right)$$

                  $$ \Rightarrow a\left( {{b^2} - ac} \right) - 2b\left( {3b - 4c} \right) + 2c\left( {3a - 4b} \right) = 0$$              $$\left( \mathbf{\because A = 0} \right)$$

                  $$ \Rightarrow a{b^2} - {a^2}c - 6{b^2} + 8bc + 6ac - 8bc = 0$$

                  $$ \Rightarrow a{b^2} - 6{b^2} = {a^2}c - 6ac$$

                  $$ \Rightarrow \left( {a - 6} \right){b^2} = ac\left( {a - 6} \right)$$

                  $$ \Rightarrow {b^2} = ac$$

                  $$ \Rightarrow {b^3} = abc$$     $$\left( {{\textbf{multiplying both side by b}}} \right)$$  

$${\textbf{ Hence, abc = }}{{\textbf{b}}^\mathbf 3.}$$ $$\textbf{The correct option is C.}$$

$$\begin{vmatrix} 1 & cos\alpha  & cos\beta  \\ cos\alpha  & 1 & cos\gamma  \\ cos\beta  & cos\gamma  & 1 \end{vmatrix}=\begin{vmatrix} 0 & cos\alpha  & cos\beta  \\ cos\alpha  & 0 & cos\gamma  \\ cos\beta  & cos\gamma  & 0 \end{vmatrix}$$

$$ \Rightarrow 1-\cos ^{ 2 }{ \gamma - } \cos { \alpha  } (\cos { \alpha  } -\cos { \beta  } \cos { \gamma  } )+\cos { \beta  } (\cos { \alpha  } \cos { \gamma  } -\cos { \beta  } )=-cos\alpha (-\cos { \beta  } \cos { \gamma  } )+cos\beta (cos\alpha cos\gamma )$$

$$ \Rightarrow 1-\cos ^{ 2 }{ \gamma - } \cos ^{ 2 }{ \alpha + } \cos { \alpha \cos { \beta \cos { \gamma  }  } + } \cos { \alpha \cos { \beta \cos { \gamma  }  } - } \cos ^{ 2 }{ \beta  } =\cos { \alpha \cos { \beta \cos { \gamma  }  } + } \cos { \alpha \cos { \beta \cos { \gamma  }  }  } $$

$$ \Rightarrow \cos ^{ 2 }{ \alpha + } \cos ^{ 2 }{ \beta  } +\cos ^{ 2 }{ \gamma =1 } $$

$$|f(x)|=\begin{bmatrix} \sin { x }  & \csc { x }  & \tan { x }  \\ \sec { x }  & x\sin { x }  & x\tan { x }  \\ x^{ 2 }-1 & -\sin { x }  & x^{ 2 }+1 \end{bmatrix}$$

$$\int _{ -a }^{ a }{ |f(x)|dx= } \begin{bmatrix} \int _{ -a }^{ a }{ \sin { x } dx }  & \int _{ -a }^{ a }{ \csc { x } dx }  & \int _{ -a }^{ a }{ \tan { x } dx }  \\ \int _{ -a }^{ a }{ \sec { x } dx }  & \int _{ -a }^{ a }{ x\sin { x } dx }  & \int _{ -a }^{ a }{ x\tan { x } dx }  \\ \int _{ -a }^{ a }{ (x^{ 2 }-1)dx }  & -\int _{ -a }^{ a }{ \sin { x } dx }  & \int _{ -a }^{ a }{ (x^{ 2 }+1)dx }  \end{bmatrix}$$

Co-factor of the matrix is given by, 

The given matrix can be solved as further,

Cofactor of the matrix is given by 

$$\displaystyle A= \begin{bmatrix} 1 & 5 & -6 \\ -8 & 0 & 4 \\ 3 & -7 & 2 \end{bmatrix}$$