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

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

Answers (1)

Answer:

\begin{aligned} &x=\frac{1}{2} \quad, \quad y=\frac{1}{3} \quad, \quad z=\frac{1}{5} \end{aligned}

Given:

\begin{aligned} &\frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10 \\ &\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10 \\ &\frac{3}{x}-\frac{1}{y}+\frac{3}{z}=13 \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:

\text { Let } \frac{1}{x} \text { be } a, \frac{1}{y} \text { be } b, \frac{1}{z} \text { be } c

        \begin{aligned} &A=\left[\begin{array}{ccc} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{array}\right] \\ &{\left[\begin{array}{ccc} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 14 \\ 4 \\ 0 \end{array}\right]} \end{aligned}

        A\; X=B\\ \begin{aligned} |A|=\left|\begin{array}{ccc} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{array}\right| &=2(2+1)+3(2-3)+3(-1-3) \\ &=6-3-12 \\ &=-9 \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} 1 & 1 \\ -1 & 2 \end{array}\right|=3 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 1 & 1 \\ 3 & 2 \end{array}\right|=1 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 1 & 1 \\ 3 & -1 \end{array}\right|=-4 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} -3 & 3 \\ -1 & 2 \end{array}\right|=3 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 2 & 3 \\ 3 & 2 \end{array}\right|=-5 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 2 & -3 \\ 3 & -1 \end{array}\right|=-7 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -3 & 3 \\ 1 & 1 \end{array}\right|=-6 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 3 \\ 1 & 1 \end{array}\right|=1 \end{aligned}

        \begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 2 & -3 \\ 1 & 1 \end{array}\right|=5 \\ \operatorname{adjA} &=\left[\begin{array}{ccc} 3 & 1 & -4 \\ 3 & -5 & -7 \\ -6 & 1 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{array}\right] \\ X=& A^{-1} B \end{aligned}

        \begin{aligned} &=-\frac{1}{9}\left[\begin{array}{ccc} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 10 \\ 13 \end{array}\right] \\ &=-\frac{1}{9}\left[\begin{array}{c} 30+30-78 \\ 10-50+13 \\ -40-70+65 \end{array}\right] \\ &=-\frac{1}{9}\left[\begin{array}{r} -18 \\ -27 \\ -45 \end{array}\right] \end{aligned}

        \begin{aligned} &\frac{1}{x}=a=\frac{-9}{-18}, \frac{1}{y}=b=\frac{-9}{-27}, \frac{1}{z}=c=\frac{-9}{-45} \\ &x=\frac{1}{2} \quad, \quad y=\frac{1}{3} \quad, \quad z=\frac{1}{5} \end{aligned}

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Gurleen Kaur

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