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

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

Answers (1)

Answer:

\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}

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