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  • Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 7 maths textbook solution

Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 7 maths textbook solution

Answers (1)

Answer:

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

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right] \\ &x+2 y+5 z=10 \\ &x-y-z=-2 \\ &2 x+3 y-z=-11 \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:

        \begin{aligned} A &=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right] \\ |A| &=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right| \\ |A| &=1(1+3)-2(-1+2)+5(3+2) \\ &=4-2+25=27 \end{aligned}

        C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,

        \begin{array}{ll} C_{11}=-1^{1+1}\left|\begin{array}{cc} -1 & -1 \\ 3 & -1 \end{array}\right|=4 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 2 & 5 \\ 3 & -1 \end{array}\right|=-1 \\ C_{12}=-1^{1+2}\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|=-1 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 1 & 5 \\ 2 & -1 \end{array}\right|=17 \end{array}

        \begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right|=5 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|=1 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} 2 & 5 \\ -1 & -1 \end{array}\right|=3 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & 5 \\ 1 & -1 \end{array}\right|=6 \end{aligned}

        \begin{aligned} &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & 2 \\ 1 & -1 \end{array}\right|=-3 \\ &\operatorname{adjA}=\left[\begin{array}{ccc} 4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3 \end{array}\right]^{T}=\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}

        \begin{gathered} =\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \\ {\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10 \\ -2 \\ -11 \end{array}\right]} \\ A X=B \end{gathered}

        \begin{aligned} &X=A^{-1} B \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right]\left[\begin{array}{c} 10 \\ -2 \\ -11 \end{array}\right]} \end{aligned}

        \begin{aligned} &\frac{1}{27}\left[\begin{array}{c} 40-34-33 \\ -10+22-66 \\ 50-2+33 \end{array}\right] \\ &=\frac{1}{27}\left[\begin{array}{c} -27 \\ -54 \\ 81 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \frac{-27}{27} \\ \frac{-54}{27} \\ \frac{81}{27} \end{array}\right]=\left[\begin{array}{c} -1 \\ -2 \\ 3 \end{array}\right]} \end{aligned}

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

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Gurleen Kaur

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