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

Explain solution RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (ii) maths

Answers (1)

Answer:

x=-\frac{8}{7}\; \; ,\; \; y=\frac{10}{7}\; \; ,\; \; z=\frac{19}{7}

Given:

\begin{aligned} &x+y+z=3 \\ &2 x-y+z=-1 \\ &2 x+y-3 z=-9 \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 .

Solution:

        \begin{aligned} &{\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 3 \\ -1 \\ -9 \end{array}\right]} \\ &A X=B \end{aligned}

        \begin{aligned} |A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3 \end{array}\right| &=1(3-1)-1(6-2)+1(2+2) \\ &=2+8+4 \\ &=14 \neq 0 \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} -1 & 1 \\ 1 & -3 \end{array}\right|=2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ 2 & -3 \end{array}\right|=8 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & -1 \\ 2 & 1 \end{array}\right|=4 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ 1 & -3 \end{array}\right|=4 \end{aligned}

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

        \begin{aligned} &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right|=-3 \\ &\operatorname{adj} A=\left[\begin{array}{ccc} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{array}\right]^{T} \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}

        \begin{aligned} &=\frac{1}{14}\left[\begin{array}{ccc} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{array}\right] \\ X=& A^{-1} B \end{aligned}

        \begin{aligned} &=\frac{1}{14}\left[\begin{array}{ccc} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \\ 9 \end{array}\right] \\ &=\frac{1}{14}\left[\begin{array}{c} 6-4-18 \\ 24+5-9 \\ 12-1+27 \end{array}\right] \\ &=\frac{1}{14}\left[\begin{array}{c} -16 \\ 20 \\ 38 \end{array}\right] \end{aligned}

        x=-\frac{16}{14}\; \; ,\; \; y=\frac{20}{14}\; \; ,\; \; z=\frac{38}{14} \\ x=-\frac{8}{7}\; \; ,\; \; y=\frac{10}{7}\; \; ,\; \; z=\frac{19}{7}

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Gurleen Kaur

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