Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 15

\begin{aligned} x=4000, y=5000, z=3000 \end{aligned}

Given:

According to question

\begin{aligned} &4 x+3 y+2 z=37000 \\ &5 x+3 y+4 z=47000 \\ &x+y+z=12000 \end{aligned}

Solution:

\begin{aligned} &{\left[\begin{array}{lll} 4 & 3 & 2 \\ 5 & 3 & 4 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 37000 \\ 47000 \\ 12000 \end{array}\right]} \\ &A X=B \\ &|A|=4(3-4)-3(5-4)+2(5-3) \\ &=-4-3+4=-3 \neq 0 \end{aligned}

$So, A\: is\; non\; singular \\ \therefore inverse\; exists \\ A_{11}=-1 A_{12}=-1 A_{13}=2 \\ A_{21}=-1 A_{22}=2 A_{23}=-1 \\ A_{31}=6 A_{32}=-6 A_{33}=-3 \\ \operatorname{adj} A=\left[\begin{array}{ccc}-1 & -1 & 6 \\ -1 & 2 & -6 \\ 2 & -1 & -3\end{array}\right]$

\begin{aligned} &A^{-1}=\frac{1}{|A|} \operatorname{adj} A=-\frac{1}{3}\left[\begin{array}{ccc} -1 & -1 & 6 \\ -1 & 2 & -6 \\ 2 & -1 & -3 \end{array}\right] \\ &X=A^{-1} B=-\frac{1}{3}\left[\begin{array}{ccc} -1 & -1 & 6 \\ -1 & 2 & -6 \\ 2 & -1 & -3 \end{array}\right] \\ &=\frac{1}{6}\left[\begin{array}{c} -37000-47000+72000 \\ -37000+97000-72000 \\ 74000-47000-36000 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4000 \\ 5000 \\ 3000 \end{array}\right]} \\ &x=4000, y=5000, z=3000 \end{aligned}