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

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

Answers (1)

Answer:

\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}

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Gurleen Kaur

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