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

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

Answers (1)

Answer:

\begin{aligned} &x=1 \quad, \quad y=2 \quad, \quad z=5 \end{aligned}

Given:

\begin{aligned} &5 x+3 y+z=16 \\ &2 x+y+3 z=19 \\ &x+2 y+4 z=25 \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}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right] \\ &\begin{aligned} |A|=\left|\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right|=5(4-6)-3(8-3)+1(4-1) \\ \end{aligned} \end{aligned}

                                       =-10-15+3 \\ =-22

        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}{ll} 1 & 3 \\ 2 & 4 \end{array}\right|=-2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right|=-5 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right|=-10 \end{aligned}

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 5 & 1 \\ 1 & 4 \end{array}\right|=19 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=-7 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|=8 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 5 & 1 \\ 2 & 3 \end{array}\right|=-13 \end{aligned}

        \begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 5 & 3 \\ 2 & 1 \end{array}\right|=-1 \\ \operatorname{adj} A &=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=-\frac{1}{22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right] \\ &=-\frac{1}{22}\left[\begin{array}{c} -32-190+200 \\ -80+361-325 \\ 48-133-25 \end{array}\right] \end{aligned}

        \begin{aligned} &=-\frac{1}{22}\left[\begin{array}{c} -22 \\ -44 \\ -110 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right]} \\ &x=1, \quad y=2 \quad, \quad z=5 \end{aligned}

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Gurleen Kaur

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