Maths

Kindly go through the solution

Kindly go through the solution

A $\left|\begin{array}{ccc}\hfill x+2\hfill & \hfill x+3\hfill & \hfill x+2a\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+2b\hfill \\ \hfill x+4\hfill & \hfill x+5\hfill & \hfill x+2c\hfill \end{array}\right|=0$

Given equations are :-

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

This system of equation can be written, in matrix form, as AX= B, Where

⇒ x = 2, y = 3, z = 5.

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill sin\alpha \hfill & \hfill cos\alpha \hfill & \hfill cos\left(\alpha +\delta \right)\hfill \\ \hfill sin\beta \hfill & \hfill cos\beta \hfill & \hfill cos\left(\beta +\delta \right)\hfill \\ \hfill sin\gamma \hfill & \hfill cos\gamma \hfill & \hfill cos\left(\gamma +\delta \right)\hfill \end{array}\right|\\ =\frac{1}{sin\delta cos\delta }\left|\begin{array}{ccc}\hfill sin\alpha sin\delta \hfill & \hfill cos\alpha cos\delta \hfill & \hfill cos\left(\alpha +\delta \right)\hfill \\ \hfill sin\beta sin\delta \hfill & \hfill cos\beta cos\delta \hfill & \hfill cos\left(\beta +\delta \right)\hfill \\ \hfill sin\gamma sin\delta \hfill & \hfill cos\gamma cos\delta \hfill & \hfill cos\left(\gamma +\delta \right)\hfill \end{array}\right|\end{array}$

Applying $cos\left(A+B\right)=cosAcosB-sinAsinB$ in $c_3$

$\begin{array}{l}{c}_1\to c_1+c_3\\ ?=\frac{1}{sin\delta cos\delta }\left|\begin{array}{ccc}\hfill cos\alpha cos\delta \hfill & \hfill cos\alpha cos\delta \hfill & \hfill cos\alpha cos\delta -sin\alpha sin\delta \hfill \\ \hfill cos\beta cos\delta \hfill & \hfill cos\beta cos\delta \hfill & \hfill cos\beta cos\delta -sin\beta sin\delta \hfill \\ \hfill cos\gamma cos\delta \hfill & \hfill cos\gamma cos\delta \hfill & \hfill cos\gamma cos\delta -sin\gamma sin\delta \hfill \end{array}\right|\\ c_1=c_2\\ \therefore ?=0=R.H.S.\end{array}$

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 2\hfill & \hfill 3+2p\hfill & \hfill 4+3p+2q\hfill \\ \hfill 3\hfill & \hfill 6+3p\hfill & \hfill 10+6p+3q\hfill \end{array}\right|\\ R_2\to R_2-2R_1\\ R_3\to R_3-3R_1\\ =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 7+3p\hfill \end{array}\right|\\ R_3\to R_3-3R_2\\ =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

Expanding along $c_1$

$?=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=1=R.H.S.$

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill 3a\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill -b+a\hfill & \hfill 3b\hfill & \hfill -b+c\hfill \\ \hfill -c+a\hfill & \hfill -c+b\hfill & \hfill 3c\hfill \end{array}\right|\\ c_1\to c_1+c_2+c_3\\ =\left|\begin{array}{ccc}\hfill a+b+c\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill a+b+c\hfill & \hfill 3b\hfill & \hfill -b+c\hfill \\ \hfill a+b+c\hfill & \hfill -c+b\hfill & \hfill 3c\hfill \end{array}\right|\\ \left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill 1\hfill & \hfill 3b\hfill & \hfill -b+c\hfill \\ \hfill 1\hfill & \hfill -c+b\hfill & \hfill 3c\hfill \end{array}\right|\end{array}$

$R_2\to R_2-R_1$and $R_3\to R_3-R_1$

$=\left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill 0\hfill & \hfill 3b+a-b\hfill & \hfill -b+a\hfill \\ \hfill 0\hfill & \hfill -a+b+a-b\hfill & \hfill 3c+a-c\hfill \end{array}\right|$

Expanding along ${}_{}$$c_1$

$\begin{array}{l}=\left(a+b+c\right)\left|\begin{array}{cc}\hfill 2b+a\hfill & \hfill a-b\hfill \\ \hfill a-c\hfill & \hfill 2c+a\hfill \end{array}\right|\\ =\left(a+b+c\right)\left[4bc+2ab+2ac+a^2+ac+ab-bc\right]\\ =\left(a+b+c\right)\left(3ab+3bc+3ac\right)\\ =3\left(a+b+c\right)\left(ab+bc+ac\right)\\ =R.H.S.\end{array}$

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1+p{x}^3\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1+p{y}^3\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1+p{z}^3\hfill \end{array}\right|\\ =\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill p{x}^3\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill p{y}^3\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill p{z}^3\hfill \end{array}\right|\\ ={?}_1+{?}_2-----(1)\\ Now{?}_2=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill p{x}^3\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill p{y}^3\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill p{z}^3\hfill \end{array}\right|\\ =pxyz\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill x^2\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill y^2\hfill \\ \hfill 1\hfill & \hfill z\hfill & \hfill z^2\hfill \end{array}\right|\\ c_1\leftrightarrow c_3\\ =-pxyz\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x\hfill & \hfill 1\hfill \\ \hfill y^2\hfill & \hfill y\hfill & \hfill 1\hfill \\ \hfill z^2\hfill & \hfill z\hfill & \hfill 1\hfill \end{array}\right|\\ c_1\leftrightarrow c_2\\ =pxyz\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1\hfill \end{array}\right|=pxyz{?}_1\\ \end{array}$

Putting value in (1)

$\begin{array}{l}\Rightarrow {?}_1+pxyz{?}_1\\ =\left(1+pxyz\right){?}_1-----(2)\\ Now{?}_1=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y-x\hfill & \hfill y^2-x^2\hfill & \hfill 0\hfill \\ \hfill z-x\hfill & \hfill z^2-x^2\hfill & \hfill 0\hfill \end{array}\right|$

Expanding along $c_3$

$\begin{array}{l}{?}_1=\left|\begin{array}{cc}\hfill \left(y-x\right)\hfill & \hfill \left(y-x\right)\left(y+x\right)\hfill \\ \hfill \left(z-x\right)\hfill & \hfill \left(z-x\right)\left(z+x\right)\hfill \end{array}\right|\\ =\left(y-x\right)\left(z-x\right)\left|\begin{array}{cc}\hfill 1\hfill & \hfill y+x\hfill \\ \hfill 1\hfill & \hfill z+x\hfill \end{array}\right|\\ =\left(y-x\right)\left(z-x\right)\left(z+x-y-x\right)\\ =\left(x-y\right)\left(y-z\right)\left(z-x\right)\end{array}$

Putting ${?}_1$ in (2)

$L.H.S.=\left(1+pxyz\right)\left(x-y\right)\left(y-z\right)\left(z-x\right)=R.H.S$

  $\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill \beta +\gamma \hfill \\ \hfill \beta \hfill & \hfill {\beta }^2\hfill & \hfill \gamma +\alpha \hfill \\ \hfill \gamma \hfill & \hfill {\gamma }^2\hfill & \hfill \alpha +\beta \hfill \end{array}\right|\\ c_3\to c_3+c_1\\ =\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill \alpha +\beta +\gamma \hfill \\ \hfill \beta \hfill & \hfill {\beta }^2\hfill & \hfill \alpha +\beta +\gamma \hfill \\ \hfill \gamma \hfill & \hfill {\gamma }^2\hfill & \hfill \alpha +\beta +\gamma \hfill \end{array}\right|\\ =\left(\alpha +\beta +\gamma \right)\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill 1\hfill \\ \hfill \beta \hfill & \hfill {\beta }^2\hfill & \hfill 1\hfill \\ \hfill \gamma \hfill & \hfill {\gamma }^2\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$=\left(\alpha +\beta +\gamma \right)\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill 1\hfill \\ \hfill \beta -\alpha \hfill & \hfill {\beta }^2-{\alpha }^2\hfill & \hfill 0\hfill \\ \hfill \gamma -\alpha \hfill & \hfill {\gamma }^2-{\alpha }^2\hfill & \hfill 0\hfill \end{array}\right|$

Expanding along $c_3$

$\begin{array}{l}=\left(\alpha +\beta +\gamma \right)\left|\begin{array}{cc}\hfill \beta -\alpha \hfill & \hfill {\beta }^2-{\alpha }^2\hfill \\ \hfill \gamma -\alpha \hfill & \hfill {\gamma }^2-{\alpha }^2\hfill \end{array}\right|\\ =\left(\alpha +\beta +\gamma \right)\left|\begin{array}{cc}\hfill \beta -\alpha \hfill & \hfill \left(\beta -\alpha \right)\left(\beta +\alpha \right)\hfill \\ \hfill \gamma -\alpha \hfill & \hfill \left(\gamma -\alpha \right)\left(\gamma +\alpha \right)\hfill \end{array}\right|\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left|\begin{array}{cc}\hfill 1\hfill & \hfill \beta +\gamma \hfill \\ \hfill 1\hfill & \hfill \gamma +\alpha \hfill \end{array}\right|\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left\{\gamma +2-\left(\beta -\gamma \right)\right\}\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left(\gamma -\beta \right)\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\alpha -\beta \right)\left(\beta -\gamma \right)\left(\gamma -\alpha \right)=R.H.S.\end{array}$