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  • Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xi) maths textbook solution

Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xi) maths textbook solution

Answers (1)

Answer:

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

Given:

\begin{aligned} &8 x+4 y+3 z=18 \\ &2 x+y+z=5 \\ &x+2 y+z=5 \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}{lll} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 18 \\ 5 \\ 5 \end{array}\right]} \\ &A X=B \end{aligned}

        \begin{aligned} |A|=\left|\begin{array}{lll} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right|=& 8(1-2)-4(2-1)+3(4-1) \\ &=-8-4+9 \\ &=-3 \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}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=-1 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}\right|=-1 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 4 & 3 \\ 2 & 1 \end{array}\right|=2 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 8 & 3 \\ 1 & 1 \end{array}\right|=5 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 8 & 4 \\ 1 & 2 \end{array}\right|=-12 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 4 & 3 \\ 1 & 1 \end{array}\right|=1 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 8 & 3 \\ 2 & 1 \end{array}\right|=-2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 8 & 4 \\ 2 & 1 \end{array}\right|=0 \end{aligned}

        \begin{aligned} \text { adjA } &=\left[\begin{array}{ccc} -1 & -1 & 3 \\ 2 & 5 & -12 \\ -1 & -2 & 0 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=-\frac{1}{3}\left[\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=-\frac{1}{3}\left[\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0 \end{array}\right]\left[\begin{array}{c} 18 \\ 5 \\ 5 \end{array}\right] \\ &=-\frac{1}{3}\left[\begin{array}{c} -18+10+5 \\ -18+25-10 \\ 54-60 \end{array}\right] \end{aligned}

        \begin{aligned} &=-\frac{1}{3}\left[\begin{array}{r} -3 \\ -3 \\ -6 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]} \\ &x=1 \quad, \quad y=1 \quad, \quad z=2 \end{aligned}

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