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

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

Answers (1)

Answer:

\begin{aligned} & x=-2 \quad, \quad \mathrm{y}=1 \quad, \quad z=2 \end{aligned}

Given:

\begin{aligned} &2 x+6 y=2 \\ &3 x-z=-8 \\ &2 x-y+z=-3 \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} &{\left[\begin{array}{ccc} 2 & 6 & 0 \\ 3 & 0 & -1 \\ 2 & -1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -8 \\ -3 \end{array}\right]} \\ &A X=B \end{aligned}

        \begin{aligned} |A|=\left|\begin{array}{ccc} 2 & 6 & 0 \\ 3 & 0 & -1 \\ 2 & -1 & 1 \end{array}\right| &=2(0-1)-6(3+2)+0(-3+0) \\ &=-2-30 \\ &=-32 \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} 0 & -1 \\ -1 & 1 \end{array}\right|=-1 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & -1 \\ 2 & 1 \end{array}\right|=-5 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 0 \\ 2 & -1 \end{array}\right|=-3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 6 & 0 \\ -1 & 1 \end{array}\right|=-6 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 2 & 0 \\ 2 & 1 \end{array}\right|=2 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 2 & 6 \\ 2 & -1 \end{array}\right|=14 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 6 & 0 \\ 0 & -1 \end{array}\right|=-6 \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 0 \\ 3 & -1 \end{array}\right|=2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 2 & 6 \\ 3 & 0 \end{array}\right|=-18 \end{aligned}

        \begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} -1 & -5 & -3 \\ -1 & 2 & 14 \\ -6 & 2 & -18 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=-\frac{1}{32}\left[\begin{array}{ccc} -1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=-\frac{1}{32}\left[\begin{array}{ccc} -1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18 \end{array}\right]\left[\begin{array}{c} 2 \\ -8 \\ -3 \end{array}\right] \\ &=-\frac{1}{32}\left[\begin{array}{c} -2+48+18 \\ -10-16-6 \\ -6-112+54 \end{array}\right] \end{aligned}

        \begin{aligned} &=-\frac{1}{32}\left[\begin{array}{c} 64 \\ -32 \\ -64 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -2 \\ 1 \\ 2 \end{array}\right]} \\ &x=-2, \quad y=1 \quad, \quad z=2 \end{aligned}

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Gurleen Kaur

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