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

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

Answers (1)

Answer:

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

Given:

\begin{aligned} &x-y+z=2 \\ &2 x-y=0 \\ &2 y-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 using Adj A calculate A-1.

Solution:

        \begin{array}{r} {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right]} \end{array}\\ \ A\; X=B

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

        \begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right|=-1 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right|=-2 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} -1 & 1 \\ -1 & 0 \end{array}\right|=1 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 2 & 0 \end{array}\right|=2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|=1 \end{aligned}

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

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ 4 & -2 & 1 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=\frac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ 4 & -2 & 1 \end{array}\right]\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right] \\ &=\frac{1}{3}\left[\begin{array}{l} 2+1 \\ 4+2 \\ 8+1 \end{array}\right] \end{aligned}

                \begin{aligned} =\frac{1}{3}\left[\begin{array}{l} 3 \\ 6 \\ 9 \end{array}\right] \end{aligned}

        \begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ &x=1 \quad, \quad y=2 \quad, \quad z=3 \end{aligned}

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Gurleen Kaur

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