Class 12th

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 $R_1$

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

Multiplying by -2

$\begin{array}{l}\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\ \Rightarrow a^2+a^2+b^2+b^2+c^2+c^2--2ab-2bc-2ca=0\\ \Rightarrow \left(a^2+b^2-2ab\right)+\left(a^2+c^2-2ac\right)+\left(b^2+c^2-2bc\right)=0\\ \Rightarrow {\left(a-b\right)}^2+{\left(a-c\right)}^2+{\left(b-c\right)}^2=0\\ \Rightarrow a-b=0,b-c=0,c-a=0\\ \left[?x^2+y^2+z^2=0\Rightarrow x=0,y=0,z=0\right]\\ \Rightarrow a=b,b=c,c=a\\ \Rightarrow a=b=c\end{array}$

$\therefore$ Either a+b+c=0 or a=b=c

$=-sin\alpha \left(-si{n}^2\beta sin\alpha -sin\alpha co{s}^2\beta \right)-0\left(cos\alpha cos\beta sin\alpha sin\beta -cos\alpha sin\beta sin\alpha cos\beta \right)+cos\alpha \left(cos\alpha co{s}^2\beta +cos\alpha si{n}^2\beta \right)$

$=si{n}^2\alpha \left(si{n}^2\beta +co{s}^2\beta \right)+co{s}^2\alpha \left(co{s}^2\beta +si{n}^2\beta \right)$

$=si{n}^2\alpha +co{s}^2\alpha$

=1

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

Multiplying R₁, by a, R₂ by b and R₃ by c

$=\frac{1}{abc}\left|\begin{array}{ccc}\hfill a^2\hfill & \hfill a^3\hfill & \hfill abc\hfill \\ \hfill b^2\hfill & \hfill b^3\hfill & \hfill abc\hfill \\ \hfill c^2\hfill & \hfill c^3\hfill & \hfill abc\hfill \end{array}\right|$

Taking abc common from c_{3 = $=\frac{abc}{abc}\left|\begin{array}{ccc}\hfill a^2\hfill & \hfill a^3\hfill & \hfill 1\hfill \\ \hfill b^2\hfill & \hfill b^3\hfill & \hfill 1\hfill \\ \hfill c^2\hfill & \hfill c^3\hfill & \hfill 1\hfill \end{array}\right|$}

Inter changing c₁ and c_{3 = $=-\left|\begin{array}{ccc}\hfill 1\hfill & \hfill a^3\hfill & \hfill a^2\hfill \\ \hfill 1\hfill & \hfill b^3\hfill & \hfill b^2\hfill \\ \hfill 1\hfill & \hfill c^3\hfill & \hfill c^2\hfill \end{array}\right|$}

Inter changing c_{2 and} c_{3 = $=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill a^2\hfill & \hfill a^3\hfill \\ \hfill 1\hfill & \hfill b^2\hfill & \hfill b^3\hfill \\ \hfill 1\hfill & \hfill c^2\hfill & \hfill c^3\hfill \end{array}\right|=R.H.S$}

Kindly go through the solution

Let k, y and z be the cost per kg of onion, wheat and rice. Then, we form a system of equations as follows

         4k + 3y + 2z = 60

         2k + 4y + 6z = 90

         6k + 2y + 3z = 70

In matrix form we can write,

AX = B

= 0 - 3 (-30) + 2 (-20)

= 90 - 40

= 50 ≠ 0.

∴ The system has a unique solution.

The cost of onion, wheat and rice per kg are? 5? 8 and? 8 respectively.