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  • Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (vii)

Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (vii)

Answers (1)

Answer:

\begin{aligned} &x=0, y=5, z=3 \end{aligned}

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] \\ &x+3 z=9 \\ &-x+2 y-2 z=4 \\ &2 x-3 y+4 z=-z \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, and that will be Adj A using Adj A calculate A-1.

Solution:

        A=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right], \quad B=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]

        \begin{aligned} A B &=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] \\ &=\left[\begin{array}{ccc} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{array}\right] \\ &=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}

        Since, \\ A \times B=1 \\ \\ B=A^{-1} \\ \\ N\! ow,\: f\! r\! om\: given\: equation \\ \\ \left[\begin{array}{ccc}1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right] \\ A^{T}=\left[\begin{array}{ccc}1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4\end{array}\right]

        So, \\\begin{aligned} \qquad \begin{array}{l} A^{T}\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left(A^{T}\right)^{-1}\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=B^{T}\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right]} \end{array} \end{aligned}

        \begin{aligned} &A^{T}\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left(A^{T}\right)^{-1}\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=B^{T}\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right]} \end{aligned}

        \begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]^{T}\left[\begin{array}{c} 9 \\ 4 \\ -3 \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -18+36-18 \\ 0+8-3 \\ 9-12+6 \end{array}\right]=\left[\begin{array}{l} 0 \\ 5 \\ 3 \end{array}\right]} \\ &\text { Hence, } x=0, y=5, z=3 \end{aligned}

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Gurleen Kaur

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