#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 18

\begin{aligned} x=100, y=200, z=300 \end{aligned}

Given:

According to question

\begin{aligned} &x+y+z=600 \\ &3 x+2 y+z=1000 \\ &4 x+y+3 z=1500 \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:

Let the award money given for discipline, politeness & punctuality be x, y, z respectively.

Since, the total cash award is 600

\begin{aligned} &x+y+z=600 \; \; \; \; \; \; \; ......(1)\\ \end{aligned}

Award money given by school P is 1000

\begin{aligned} 3 x+2 y+z=1000 \; \; \; \; \; \; ......(2) \end{aligned}

Award money given by school Q is 1500

\begin{aligned} &4 x+y+3 z=1500\; \; \; \; \; \; \; \; ......(3) \end{aligned}

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

$\left[\begin{array}{lll} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 600 \\ 1000 \\ 1500 \end{array}\right]$

\begin{aligned} &A X=B\\ &|A|=\left|\begin{array}{lll} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3 \end{array}\right|=1(6-1)-1(9-4)+1(3-8)\\ &=5-5-5=-5 \neq 0\\ &\text { Let } c_{i j} \text { be the cofactors, } \end{aligned}

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

\begin{aligned} &c_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 4 & 3 \end{array}\right|=-1, c_{23}=(-1)^{2+13}\left|\begin{array}{ll} 1 & 1 \\ 4 & 1 \end{array}\right|=3 \\ &c_{31}=(-1)^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=-1, c_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 3 & 1 \end{array}\right|=2 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & 1 \\ 3 & 2 \end{array}\right|=-1 \end{aligned}

\begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} 5 & -5 & -5 \\ -2 & -1 & 3 \\ -1 & 2 & -1 \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & -2 & -5 \\ -5 & -1 & 2 \\ -5 & 3 & -1 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \\ &\quad=\frac{1}{-5}\left[\begin{array}{ccc} 5 & -2 & -1 \\ -5 & -1 & 2 \\ -5 & 3 & -1 \end{array}\right] \end{aligned}

\begin{aligned} &X=A^{-1} B \\ &X=\frac{1}{-5}\left[\begin{array}{ccc} 5 & -2 & -1 \\ -5 & -1 & 2 \\ -5 & 3 & -1 \end{array}\right]\left[\begin{array}{c} 600 \\ 1000 \\ 1500 \end{array}\right] \\ &X=\frac{-1}{5}\left[\begin{array}{c} 3000-2000-1500 \\ 3000-1000+3000 \\ -3000+3000-1500 \end{array}\right] \end{aligned}

\begin{aligned} &=-\frac{1}{5}\left[\begin{array}{l} -500 \\ -1000 \\ -1500 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \frac{-500}{-5} \\ \frac{-1000}{-5} \\ \frac{-1500}{-5} \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 100 \\ 200 \\ 300 \end{array}\right]} \\ &x=100, y=200, z=300 \end{aligned}