#### Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (iii) maths textbook solution

$x=\frac{1}{2}\; \; ,\; \; y=\frac{1}{3}\; \; ,\; \; z=\frac{1}{5}$

Given:

\begin{aligned} &6x-12y+25z=4 \\ &4 x+15 y-20z=3 \\ &2 x+18y+15 z=10 \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 .

Solution:

$A=\left[\begin{array}{ccc} 6 & -12 & 25 \\ -12 & 15 & -20 \\ 2 & 18 & 15 \end{array}\right]$

$\begin{gathered} |A|=\left|\begin{array}{ccc} 6 & -12 & 25 \\ -12 & 15 & -20 \\ 2 & 18 & 15 \end{array}\right|=6(225+360)+12(60+40)+25(72-30) \\ = 3510+1200+1050 \\ =5760 \end{gathered}$

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

\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 15 & -20 \\ 18 & 15 \end{array}\right|=585 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & -20 \\ 2 & 15 \end{array}\right|=-100 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 4 & 15 \\ 2 & 18 \end{array}\right|=42 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 12 & 25 \\ 18 & 15 \end{array}\right|=630 \end{aligned}

\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 6 & 25 \\ 2 & 15 \end{array}\right|=40 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 6 & -12 \\ 2 & 8 \end{array}\right|=-132 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -12 & 25 \\ 15 & -20 \end{array}\right|=-135 \quad , \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 6 & 25 \\ 4 & -20 \end{array}\right|=220 \end{aligned}

\begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 6 & -12 \\ 4 & 15 \end{array}\right|=138 \\ \operatorname{adjA} &=\left[\begin{array}{ccc} 585 & -100 & 42 \\ 630 & 40 & -132 \\ -135 & 220 & 138 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{5760}\left[\begin{array}{ccc} 585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{5760}\left[\begin{array}{ccc} 585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \\ 9 \end{array}\right] \\ &=\frac{1}{5760}\left[\begin{array}{c} 6-4-18 \\ 24+5-9 \\ 12-1+27 \end{array}\right] \\ &=\frac{1}{5760}\left[\begin{array}{c} -16 \\ 20 \\ 38 \end{array}\right] \end{aligned}

\begin{aligned} &x=\frac{2880}{5760} \quad, \quad y=\frac{1920}{5760} \quad, \quad z=\frac{1152}{5760} \\ &x=\frac{1}{2} \quad, \quad y=\frac{1}{3} \quad, \quad z=\frac{1}{5} \end{aligned}