#### Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 12 maths

The prices of the commodities P, Q, R are x, y, z

According to Question-

\begin{aligned} &3 x+5 y-4 z=6000 \\ &2 x-3 y+z=5000 \\ &-x+4 y+6 z=13000 \end{aligned}

Hint:

X=A-1B is used to solve this problem.

And the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.

Solution:

According to the question,

\begin{aligned} &3 x+5 y-4 z=6000 \\ &2 x-3 y+z=5000 \\ &-x+4 y+6 z=13000 \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right]} \\ &A X=B \\ &|A|=\left|\begin{array}{ccc} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{array}\right|=3(-18-4)-5(12+1)-4(8-3)=-66-66-20=-151 \neq 0 \\ &\text { So } A^{-1} \text { exists } \end{aligned}

\begin{aligned} &c_{11}=(-1)^{1+1}\left|\begin{array}{cc} -3 & 1 \\ 4 & 6 \end{array}\right|=-22 \\ &c_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ -1 & 6 \end{array}\right|=-13 \\ &c_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & -3 \\ -1 & 4 \end{array}\right|=5 \\ &c_{21}=(-1)^{2+1}\left|\begin{array}{cc} 5 & -4 \\ 4 & 6 \end{array}\right|=-46 \end{aligned}

\begin{aligned} &c_{22}=(-1)^{2+2}\left|\begin{array}{cc} 3 & -4 \\ -1 & 6 \end{array}\right|=14 \\ &c_{23}=(-1)^{2+13}\left|\begin{array}{cc} 3 & 5 \\ -1 & 4 \end{array}\right|=-17 \\ &c_{31}=(-1)^{3+1}\left|\begin{array}{cc} 5 & -4 \\ -3 & 1 \end{array}\right|=-7 \\ &c_{32}=(-1)^{3+2}\left|\begin{array}{cc} 3 & -4 \\ 2 & 1 \end{array}\right|=-11 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{cc} 3 & 5 \\ 2 & -3 \end{array}\right|=-19 \end{aligned}

\begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} -22 & -13 & 5 \\ -46 & 14 & -17 \\ -7 & -11 & -19 \end{array}\right]^{T}=\left[\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \\ &=-\frac{1}{151}\left[\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right] \end{aligned}

\begin{aligned} &X=A^{-1} B \\ &X=-\frac{1}{151}\left[\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right]\left[\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right] \\ &X=\frac{-1}{151}\left[\begin{array}{l} -453000 \\ -151000 \\ -302000 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3000 \\ 1000 \\ 2000 \end{array}\right]} \\ &\therefore x=3000, y=1000, z=2000 \end{aligned}