#### Need solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 11

\begin{aligned} &x=11, y=15, z=19 \end{aligned}

Given:

Let x, y, z  be the production level of 1st,2nd,3rd product respectively.

Hint:

X = A-1is used for the problem

And the determination & cofactor of matrix A.Take transpose that will adjA using adjA find A-1

Solution:

According to the Question,

\begin{aligned} &x+y+z=45\: \: \: \: ......(1)\\ &-x+z=8\: \: \: \: ......(2)\\ &x+z=2 y\\ &x-2 y+z=0\: \: \: \: ......(3)\\ \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 45 \\ 8 \\ 0 \end{array}\right]} \\ &A X=B \end{aligned}

$|A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{array}\right|=1(-0+2)-1(-1-1)+1(2-0)=2+2+2=6$

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

\begin{aligned} &c_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & 1 \\ -2 & 1 \end{array}\right|=2 \\ &c_{12}=(-1)^{1+2}\left|\begin{array}{cc} -1 & 1 \\ 1 & 1 \end{array}\right|=2 \\ &c_{13}=(-1)^{1+3}\left|\begin{array}{cc} -1 & 0 \\ 1 & -2 \end{array}\right|=2 \\ &c_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right|=-3 \end{aligned}

\begin{aligned} &c_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|=0 \\ &c_{23}=(-1)^{2+13}\left|\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right|=3 \\ &c_{31}=(-1)^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|=1 \\ &c_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right|=-2 \\ &c_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ -1 & 0 \end{array}\right|=1 \end{aligned}

\begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} 2 & 2 & 2 \\ -3 & 0 & 3 \\ 1 & -2 & 1 \end{array}\right]^{T}=\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}

\begin{aligned} &=\frac{1}{6}\left[\begin{array}{ccc} 2 & -3 & 1 \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{array}\right] \\ X &=A^{-1} B \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 90-24+0 \\ 90+0+0 \\ 90+24+0 \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 66 \\ 90 \\ 114 \end{array}\right]=\left[\begin{array}{c} 11 \\ 15 \\ 19 \end{array}\right]} \\ &x=11, y=15, z=19 \end{aligned}