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

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

Answers (1)

The prices of the commodities P, Q, R are x, y, z

According to Question-

\begin{aligned} &3 x+5 y-4 z=6000 \\ &2 x-3 y+z=5000 \\ &-x+4 y+6 z=13000 \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, and that will be Adj A using Adj A calculate A-1.

Solution:

According to the question,

\begin{aligned} &3 x+5 y-4 z=6000 \\ &2 x-3 y+z=5000 \\ &-x+4 y+6 z=13000 \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right]} \\ &A X=B \\ &|A|=\left|\begin{array}{ccc} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{array}\right|=3(-18-4)-5(12+1)-4(8-3)=-66-66-20=-151 \neq 0 \\ &\text { So } A^{-1} \text { exists } \end{aligned}

\begin{aligned} &c_{11}=(-1)^{1+1}\left|\begin{array}{cc} -3 & 1 \\ 4 & 6 \end{array}\right|=-22 \\ &c_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ -1 & 6 \end{array}\right|=-13 \\ &c_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & -3 \\ -1 & 4 \end{array}\right|=5 \\ &c_{21}=(-1)^{2+1}\left|\begin{array}{cc} 5 & -4 \\ 4 & 6 \end{array}\right|=-46 \end{aligned}

\begin{aligned} &c_{22}=(-1)^{2+2}\left|\begin{array}{cc} 3 & -4 \\ -1 & 6 \end{array}\right|=14 \\ &c_{23}=(-1)^{2+13}\left|\begin{array}{cc} 3 & 5 \\ -1 & 4 \end{array}\right|=-17 \\ &c_{31}=(-1)^{3+1}\left|\begin{array}{cc} 5 & -4 \\ -3 & 1 \end{array}\right|=-7 \\ &c_{32}=(-1)^{3+2}\left|\begin{array}{cc} 3 & -4 \\ 2 & 1 \end{array}\right|=-11 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{cc} 3 & 5 \\ 2 & -3 \end{array}\right|=-19 \end{aligned}

\begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} -22 & -13 & 5 \\ -46 & 14 & -17 \\ -7 & -11 & -19 \end{array}\right]^{T}=\left[\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \\ &=-\frac{1}{151}\left[\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right] \end{aligned}

\begin{aligned} &X=A^{-1} B \\ &X=-\frac{1}{151}\left[\begin{array}{ccc} -22 & -46 & -7 \\ -13 & 14 & -11 \\ 5 & -17 & -19 \end{array}\right]\left[\begin{array}{c} 6000 \\ 5000 \\ 13000 \end{array}\right] \\ &X=\frac{-1}{151}\left[\begin{array}{l} -453000 \\ -151000 \\ -302000 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3000 \\ 1000 \\ 2000 \end{array}\right]} \\ &\therefore x=3000, y=1000, z=2000 \end{aligned}

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Gurleen Kaur

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