Ncert Solutions Maths class 12th

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

$R_2\to R_2-R_1$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 $c_3$

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

Kindly go through the solution

Expanding along C₁

$=2\left(x+y\right)\left\{\left(-x\right)\left(x-y\right)-y.y\right\}$

$=2\left(x+y\right)\left(-x^2+xy-y^2\right)$

$=-2\left(x+y\right)\left(x^2-xy+y^2\right)$

$=-2\left\{x^3-x^{2}y+x{y}^2+x^{2}y-x{y}^2+y^3\right\}$

= $-2\left(x^3+y^3\right)$

Kindly go through the solution

Kindly go through the solution

$L.H.S=\left|\begin{array}{ccc}\hfill a^2\hfill & \hfill bc\hfill & \hfill ac+c^2\hfill \\ \hfill a^2+ab\hfill & \hfill b^2\hfill & \hfill ac\hfill \\ \hfill ab\hfill & \hfill b^2+bc\hfill & \hfill c^2\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill a^2\hfill & \hfill bc\hfill & \hfill c\left(a+c\right)\hfill \\ \hfill a\left(a+b\right)\hfill & \hfill b^2\hfill & \hfill ac\hfill \\ \hfill ab\hfill & \hfill b\left(b+c\right)\hfill & \hfill c^2\hfill \end{array}\right|$

Taking a, b, c common from c₁, c₂ and c₃ respectively

$\begin{array}{l}=abc\left|\begin{array}{ccc}\hfill a\hfill & \hfill c\hfill & \hfill a+c\hfill \\ \hfill a+b\hfill & \hfill b\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill b+c\hfill & \hfill c\hfill \end{array}\right|\\ R_1\to R_1-R_2-R_3\\ =abc\left|\begin{array}{ccc}\hfill -2b\hfill & \hfill -2b\hfill & \hfill 0\hfill \\ \hfill a+b\hfill & \hfill b\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill b+c\hfill & \hfill c\hfill \end{array}\right|\\ c_2\to c_2-c_1\\ =abc\left|\begin{array}{ccc}\hfill -2b\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill a+b\hfill & \hfill -a\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill c\hfill \end{array}\right|\end{array}$

Expanding along $R_1$

$abc\left(-2b\right)\left(-2ac\right)=4a^2+b^2+c^2=R.H.S.$

$\begin{array}{l}\left|\begin{array}{ccc}\hfill x+a\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x+a\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x\hfill & \hfill x+a\hfill \end{array}\right|=0\\ R_1\to R_1+R_2+R_3\\ \left|\begin{array}{ccc}\hfill 3x+a\hfill & \hfill 3x+a\hfill & \hfill 3x+a\hfill \\ \hfill x\hfill & \hfill x+a\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x\hfill & \hfill x+a\hfill \end{array}\right|=0\end{array}$

Taking 3x+a common from $R_1$

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

Either $3x+a=0\Rightarrow x=-\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Or

$\begin{array}{l}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill x+a\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x\hfill & \hfill x+a\hfill \end{array}\right|=0\\ c_2\to c_2-c_1,c_3\to c_3-c_1\\ \left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill a\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill 0\hfill & \hfill a\hfill \end{array}\right|=0\end{array}$

Expanding along $R_1$ , $a^2=0$

$\Rightarrow a=0$

$\therefore$ $x=-\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ is the only solution.

$?=\left|\begin{array}{ccc}\hfill b+c\hfill & \hfill c+a\hfill & \hfill a+b\hfill \\ \hfill c+a\hfill & \hfill a+b\hfill & \hfill b+c\hfill \\ \hfill a+b\hfill & \hfill b+c\hfill & \hfill c+a\hfill \end{array}\right|=0$

$\begin{array}{l}{R}_1\to R_1+R_2+R_3\\ \left|\begin{array}{ccc}\hfill 2\left(a+b+c\right)\hfill & \hfill 2\left(a+b+c\right)\hfill & \hfill 2\left(a+b+c\right)\hfill \\ \hfill c+a\hfill & \hfill a+b\hfill & \hfill b+c\hfill \\ \hfill a+b\hfill & \hfill b+c\hfill & \hfill c+a\hfill \end{array}\right|=0\end{array}$

Taking  2(a + b + c) common from R₁

$\Rightarrow 2\left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill c+a\hfill & \hfill a+b\hfill & \hfill b+c\hfill \\ \hfill a+b\hfill & \hfill b+c\hfill & \hfill c+a\hfill \end{array}\right|=0$

Either $2\left(a+b+c\right)=0$ i.e. a+b+c=0 or

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

$c_2\to c_2-c_1$ and $c_3\to c_3-c_1$

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

Expanding along $R_1$

$\begin{array}{l}\Rightarrow \left|\begin{array}{cc}\hfill b-c\hfill & \hfill b-a\hfill \\ \hfill c-a\hfill & \hfill c-b\hfill \end{array}\right|=0\\ \Rightarrow \left(b-c\right)\left(c-b\right)-\left(b-a\right)\left(c-a\right)=0\\ \Rightarrow bc-b^2-c^2+cb-bc+ab+ac-a^2=0\\ \Rightarrow -a^2-b^2-c^2+ab+bc+ca=0\end{array}$