#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (iv)

\begin{aligned} &x=1 \quad, \quad y=1 \quad, \quad z=1 \end{aligned}

Given:

\begin{aligned} &3 x+2 y+7 z=14 \\ &2 x-y+3 z=4 \\ &x+2 y-3 z=0 \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 using Adj A calculate A-1.

Solution:

\begin{aligned} &A=\left[\begin{array}{ccc} 3 & 2 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{array}\right] \\ &{\left[\begin{array}{ccc} 3 & 2 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 14 \\ 4 \\ 0 \end{array}\right]} \end{aligned}

A\; X\; =B\\ \begin{aligned} |A|=\left|\begin{array}{ccc} 3 & 2 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{array}\right| &=3(3-6)-4(-6-3)+7(4+1) \\ &=-9+36+35 \\ &=62 \neq 0 \end{aligned}

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

$\begin{array}{ll} C_{11}=(-1)^{1+1}(3-6)=-3 & , \quad C_{12}=(-1)^{1+2}(-6-3)=9 \\ C_{13}=(-1)^{1+3}(4+1)=5 & , \quad C_{21}=(-1)^{2+1}(-12-14)=26 \end{array}$

$\begin{array}{ll} C_{22}=(-1)^{2+2}(-3-7)=-10 \quad, & C_{23}=(-1)^{2+3}(6-4)=-2 \\ C_{31}=(-1)^{3+1}(12+7)=19 & , \quad C_{32}=(-1)^{3+2}(9-14)=5 \end{array}$

\begin{aligned} C_{33}=(-1)^{3+3}(-3-8)=-11 \\ \operatorname{adjA} =\left[\begin{array}{ccc} -3 & 9 & 5 \\ 26 & -5 & -2 \\ 19 & 5 & -11 \end{array}\right]^{T} \\ =\left[\begin{array}{ccc} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{62}\left[\begin{array}{ccc} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{array}\right]\left[\begin{array}{c} 14 \\ 4 \\ 0 \end{array}\right] \\ &=\frac{1}{62}\left[\begin{array}{c} -42+104+0 \\ 126-64+0 \\ 70-8+0 \end{array}\right] \end{aligned}

\begin{aligned} &=\frac{1}{62}\left[\begin{array}{l} 62 \\ 62 \\ 62 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]} \\ &x=1 \quad, \quad y=1 \quad, \quad z=1 \end{aligned}