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  • Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (i) maths textbook solution

Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (i) maths textbook solution

Answers (1)

Answer:

x=4,\; \; y=-3,\: \: z=1

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \\ &x-2 y=10 \\ &2 x+y+3 y=8 \\ &-2 y+z=7 \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 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \\ |A| &=1(1+6)+2(2-0)+0(-4-0) \\ &=7+4+0 \\ &=11 \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 & 3 \\ -2 & 1 \end{array}\right|=7 \quad , \quad C_{21}=-1^{2+1}\left|\begin{array}{ll} -2 & 0 \\ -2 & 1 \end{array}\right|=2 \\ C_{12}=-1^{1+2}\left|\begin{array}{ll} 2 & 3 \\ 0 & 1 \end{array}\right|=-2 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \end{array}

        \begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right|=-4 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 1 & -2 \\ 0 & -2 \end{array}\right|=2 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -2 & 0 \\ 1 & 3 \end{array}\right|=-6 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & 0 \\ 2 & 3 \end{array}\right|=3 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|=5 \end{aligned}

        \begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 7 & -2 & -4 \\ 2 & 1 & 2 \\ -6 & -3 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 7 & 2 & 6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & 6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right] \end{aligned}

        \begin{aligned} X &=A^{-1} B \\ &=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & 6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ -1 \end{array}\right] \\ &=\frac{1}{11}\left[\begin{array}{c} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} 44 \\ -33 \\ 11 \end{array}\right] \end{aligned}

        \begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 4 \\ -3 \\ 1 \end{array}\right]} \\ &x=4, y=-3, z=1 \end{aligned}

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Gurleen Kaur

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