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

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

Answers (1)

Answer:

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

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & -1 \end{array}\right] \\ &2 x+y-3 z=13 \\ &3 x+2 y+z=4 \\ &x+2 y-z=8 \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} 2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & -1 \end{array}\right] \\ &\begin{aligned} |A| &=2(-2-2)-3(-1+6)+1(1+6) \\ &=-8-15+7 \\ &=-16 \neq 0 \end{aligned} \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{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{cc} 2 & 2 \\ 1 & -1 \end{array}\right|=-4 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array}\right|=-2 \\ &C_{12}=-1^{1+2}\left|\begin{array}{cc} 1 & 2 \\ -3 & -1 \end{array}\right|=-5 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 2 & 1 \\ -3 & 1 \end{array}\right|=1 \end{aligned}

        \begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 1 & 2 \\ -3 & 1 \end{array}\right|=7 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 2 & 3 \\ -3 & 1 \end{array}\right|=-11 \\ &C_{31}=-1^{3+1}\left|\begin{array}{ll} 3 & 1 \\ 2 & 2 \end{array}\right|=4 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=-3 \\ &C_{33}=-1^{3+3}\left|\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right|=3 \end{aligned}

        \begin{aligned} \text { adjA } &=\left[\begin{array}{ccc} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1 \end{array}\right] \end{aligned}

        \begin{aligned} A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{-16}\left[\begin{array}{ccc} -4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1 \end{array}\right] \end{aligned}

        \begin{aligned} &A^{T} X=B \\ &\left|A^{T}\right|=|A|=-16 \neq 0 \\ &X=(A T)^{-1} B \\ &\quad=\left(A^{-1}\right)^{T} B \end{aligned}

        \begin{aligned} &=-\frac{1}{16}\left[\begin{array}{ccc} -4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1 \end{array}\right]^{T}\left[\begin{array}{c} 13 \\ 4 \\ 8 \end{array}\right] \\ &=-\frac{1}{16}\left[\begin{array}{ccc} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{array}\right] \end{aligned}

        \begin{aligned} &=-\frac{1}{16}\left[\begin{array}{c} -52-20+56 \\ 52+4-88 \\ 52-12+8 \end{array}\right]=-\frac{1}{16}\left[\begin{array}{c} -16 \\ -32 \\ 48 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right]} \\ &x=1, y=2, z=-3 \end{aligned}

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Gurleen Kaur

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