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  • Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 6 maths


Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 6 maths

Answers (1)

Answer:

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

Given:

\begin{aligned} &2 x-3 y+5 z=11 \\ &3 x+2 y-4 z=-5 \\ &x+y+2 z=-z \\ &A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right] \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 that will be Adj A using Adj A calculate A-1.

Solution:

        \begin{aligned} &A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right] \\ &|A|=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right| \end{aligned}

        \begin{aligned} |A| &=2(-4+4)+3(-6+4)+5(3-2) \\ &=0-6+5=-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}{ll} 2 & -4 \\ 1 & -2 \end{array}\right|=0 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} -3 & 5 \\ 1 & -2 \end{array}\right|=-1 \\ &C_{12}=-1^{1+2}\left|\begin{array}{ll} 3 & -4 \\ 1 & -2 \end{array}\right|=2 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 2 & 5 \\ 1 & -2 \end{array}\right|=-9 \end{aligned}

        \begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right|=1 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 2 & -3 \\ 1 & 1 \end{array}\right|=-5 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -3 & 5 \\ 2 & -4 \end{array}\right|=2 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 2 & 5 \\ 3 & -4 \end{array}\right|=23 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 2 & -3 \\ 3 & 2 \end{array}\right|=13 \end{aligned}

        \begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{T}=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}

        \begin{gathered} =\frac{1}{-1}\left[\begin{array}{ccc} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \\ {\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 11 \\ -5 \\ -3 \end{array}\right]} \end{gathered}

        \begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 0+5-6 \\ 22+45-69 \\ 11+25-39 \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ &x=1, \mathrm{y}=2, z=3 \end{aligned}

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Gurleen Kaur

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