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  • Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xiii)

Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xiii)

Answers (1)

Answer:

\begin{aligned} &x=2 \quad, \quad y=3 \quad, \quad z=5 \end{aligned}

Given:

\begin{aligned} &\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 \\ &\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \\ &\frac{6}{x}+\frac{9}{y}-\frac{-20}{z}=2 \end{aligned}

Hint:

X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.

Solution:

        \begin{aligned} &\text { Let } \frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c\\ &\left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right]\\ &\mathrm{A} \mathrm{X}=\mathrm{B} \end{aligned}

        \begin{aligned} |A|=\left | \begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array} \right | &=2(120-45)-3(-80-30)+10(36+36) \\ &=150+330+720 \\ &=1200 \end{aligned}

        C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].

        \begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} -6 & 5 \\ 9 & -20 \end{array}\right|=75 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & 5 \\ 6 & -20 \end{array}\right|=110 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 4 & -6 \\ 6 & 9 \end{array}\right|=72 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 3 & 10 \\ 9 & -20 \end{array}\right|=150 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & 10 \\ 6 & -20 \end{array}\right|=-100 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right|=0 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 3 & 10 \\ -6 & 5 \end{array}\right|=75 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 10 \\ 4 & 5 \end{array}\right|=30 \end{aligned}

        \begin{aligned} C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 2 & 3 \\ 4 & -6 \end{array}\right|=-24 \\ \operatorname{adjA} =\left[\begin{array}{ccc} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{array}\right]^{T} \\ =\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right] \\ &=\frac{1}{1200}\left[\begin{array}{c} 300+150+150 \\ 440-100+60 \\ 288-48 \end{array}\right] \end{aligned}

        \begin{aligned} &=\frac{1}{1200}\left[\begin{array}{l} 600 \\ 400 \\ 240 \end{array}\right] \\ &\frac{1}{x}=a=\frac{1200}{600}, \frac{1}{y}=b=\frac{1200}{400}, \frac{1}{z}=c=\frac{1200}{240} \\ &x=2, \quad y=3 \quad, \quad z=5 \end{aligned}

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Gurleen Kaur

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