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

Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (i)

Answers (1)

Answer:

x=3\; \; ,\; \; y=1\; \; ,\; \; z=1

Given:

\begin{aligned} &x+y-z=3 \\ &2 x+3 y+z=10 \\ &3 x-y-7 z=1 \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} &A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{array}\right] \\ &\begin{array}{rc} |A|=\left|\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{array}\right| &=1(-2+1)-1(-14-3)-1(-2-9) \\ =-20+17+11 \\ =8 \neq 0 \end{array} \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} 3 & 1 \\ -1 & -7 \end{array}\right|=-20 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ 3 & -7 \end{array}\right|=17 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & 3 \\ 3 & -1 \end{array}\right|=-11 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 1 & -1 \\ -1 & -7 \end{array}\right|=8 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & -1 \\ 3 & -7 \end{array}\right|=-4 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 1 \\ 3 & -1 \end{array}\right|=4 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & -1 \\ 3 & 1 \end{array}\right|=4 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & -7 \\ 2 & 1 \end{array}\right|=-3 \end{aligned}

        \begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right|=1 \\ \operatorname{adj} A &=\left[\begin{array}{ccc} -20 & 17 & -11 \\ 8 & -4 & 4 \\ 4 & -3 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{8}\left[\begin{array}{ccc} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{array}\right] \\ X &=A^{-1} B \end{aligned}

        \begin{aligned} &=\frac{1}{8}\left[\begin{array}{ccc} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 3 \\ 10 \\ 1 \end{array}\right] \\ &=\frac{1}{8}\left[\begin{array}{c} -60+80+4 \\ 51-40-3 \\ -33+40+1 \end{array}\right] \\ &=\frac{1}{8}\left[\begin{array}{c} 24 \\ 8 \\ 8 \end{array}\right] \end{aligned}

        \begin{aligned} &x=\frac{24}{8} \quad, \quad \mathrm{y}=\frac{8}{8} \quad, \quad z=\frac{8}{8} \\ &\therefore x=3 \quad, \quad \mathrm{y}=1 \quad, \quad z=1 \end{aligned}

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Gurleen Kaur

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