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

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

Answers (1)

Answer:

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

Given:

\begin{aligned} &3 x+4 y+2 z=8 \\ &2 y-3 z=3 \\ &x-2 y+6 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} & A=\left[\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6 \end{array}\right] \\ &{\left[\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 8 \\ 3 \\ -2 \end{array}\right]} \end{aligned}\\ A\: \: X=B

        \begin{aligned} |A|=\left|\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6 \end{array}\right| &=3(12-6)-4(0+3)+2(0-2) \\ &=18-12-4 \\ &=2 \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} 2 & -3 \\ -2 & 6 \end{array}\right|=6 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 0 & -3 \\ 1 & 6 \end{array}\right|=-3 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 0 & 2 \\ 1 & -2 \end{array}\right|=-2 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 4 & 2 \\ -2 & 6 \end{array}\right|=-28 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 3 & 2 \\ 1 & 6 \end{array}\right|=16 \quad, \quad \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 3 & 4 \\ 1 & -2 \end{array}\right|=10 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 4 & 2 \\ 2 & -3 \end{array}\right|=-16 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 3 & 2 \\ 0 & -3 \end{array}\right|=9 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 3 & 4 \\ 0 & 2 \end{array}\right|=6 \end{aligned}

        \begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} 6 & -3 & -2 \\ -28 & 16 & 10 \\ -16 & 9 & 6 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} \operatorname{adjA} \\ &=\frac{1}{2}\left[\begin{array}{ccc} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=\frac{1}{2}\left[\begin{array}{ccc} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{array}\right]\left[\begin{array}{c} 8 \\ 3 \\ -2 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{c} 48-84+32 \\ -24+48-18 \\ -16+30-12 \end{array}\right] \end{aligned}

        \begin{aligned} &=\frac{1}{2}\left[\begin{array}{c} -4 \\ 6 \\ 2 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -2 \\ 3 \\ 1 \end{array}\right]} \\ &x=-2, \quad y=3 \quad, \quad z=1 \end{aligned}

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