#### Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 16 maths

\begin{aligned} &x=2500, y=3000, z=4000 \end{aligned}

Given:

According to question

\begin{aligned} &2 x+4 y+3 z=29000 \\ &5 x+2 y+3 z=30500 \\ &x+y+z=9500 \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 that will be Adj A using Adj A calculate A-1.

Solution:

\begin{aligned} &2 x+4 y+3 z=29000\; \; \; \; \; \; .......(1) \\ &5 x+2 y+3 z=30500\; \; \; \; \; \; .......(2) \\ &x+y+z=9500\; \; \; \; \; \; .......(3) \end{aligned}

From(1),(2),(3)

\begin{aligned} &{\left[\begin{array}{lll} 2 & 4 & 3 \\ 5 & 2 & 3 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 29000 \\ 30500 \\ 9500 \end{array}\right]} \\ &\left(R_{1} \rightarrow R_{1}-R_{3}, R_{2} \rightarrow R_{2}-5 R_{3}\right) \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 3 & 2 \\ 0 & -3 & -2 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 19500 \\ -17000 \\ 9500 \end{array}\right]} \\ &\left(R_{3} \rightarrow R_{3}-R_{1}\right) \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 3 & 2 \\ 0 & -3 & -2 \\ 0 & -2 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 19500 \\ -17000 \\ -10000 \end{array}\right]} \\ &\left(R_{3} \rightarrow 3 R_{3}-2 R_{2}\right) \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 3 & 2 \\ 0 & -3 & -2 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 19500 \\ -17000 \\ 4000 \end{array}\right]} \\ &z=4000 \\ &-3 y-2 z=-17000 \\ &-3 y-8000=-17000 \\ &-3 y=-9000 \\ &y=3000 \end{aligned}

\begin{aligned} &x+3 y+2 z=19500 \\ &x+9000+8000=19500 \\ &x=2500 \\ &\therefore x=2500, y=3000, z=4000 \end{aligned}