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  • Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (iv) maths

Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (iv) maths

Answers (1)

Answer:

x=0,\: \: y=-5,\: \: z=-3

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right] \\ &x-2 y=10 \\ &2 x-y-z=8 \\ &-2 y+z=7 \end{aligned}

Hint:

X=A-1B is used to solve this problem. First we find the determinant and 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} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right] \\ |A| &=1(-1-2)+2(2) \\ &=-3+4 \\ &=1 \end{aligned}

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

        \begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|=-3 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array}\right|=-2 \\ &C_{12}=-1^{1+2}\left|\begin{array}{cc} -2 & -2 \\ 0 & 1 \end{array}\right|=2 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \end{aligned}

        \begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} -2 & -1 \\ 0 & -1 \end{array}\right|=2 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right|=1 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right|=-4 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & 0 \\ -2 & -2 \end{array}\right|=2 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right|=3 \end{aligned}

        \begin{aligned} \operatorname{adj} A &=\left[\begin{array}{rcc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \end{aligned}

        \begin{aligned} &\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T} \\ &\text { i.e } C=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right] \\ &C^{-1}=\left[\begin{array}{rrr} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right] \end{aligned}

        \begin{gathered} {\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]} \\ C X=B \\ X=C^{-1} B \end{gathered}

        \begin{aligned} =\left[\begin{array}{ccc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right] \\ =\left[\begin{array}{c} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{array}\right] \\ x=0, y=-5, z=-3 \end{aligned}

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Gurleen Kaur

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