#### Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 9 maths textbook solution

$x=1\: \: ,\: \: y=-1\: \: ,\: \: z=2$

Given:

Sum of three numbers is 2 the second number is added to the sum of first and third, the sum is 1. By adding second and third numbers to five times the first number = 6

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:

According to question, Let the three numbers be x, y, z

\begin{aligned} &x+y+z=2 \\ &x+2 y+z=1 \\ &5 x+y+z=6 \end{aligned}

\begin{aligned} &{\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 6 \end{array}\right]} \\ &A X B \end{aligned}

\begin{aligned} |A| &=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1 \end{array}\right| \\ &=1(2-1)-1(1-5)+1(1-10) \\ &=1+4-9 \\ &=-4 \neq 0 \end{aligned}

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

\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right|=1 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|=0 \\ &C_{12}=-1^{1+2}\left|\begin{array}{ll} 1 & 1 \\ 5 & 1 \end{array}\right|=4 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 5 & 1 \end{array}\right|=-4 \end{aligned}

\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{ll} 1 & 2 \\ 5 & 1 \end{array}\right|=-9 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{ll} 1 & 1 \\ 5 & 1 \end{array}\right|=4 \\ &C_{31}=-1^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=-1 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|=0 \\ &C_{33}=-1^{3+3}\left|\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right|=1 \end{aligned}

\begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} 1 & 4 & -9 \\ 0 & -4 & 4 \\ -1 & 0 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-4}\left[\begin{array}{ccc} 1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 6 \end{array}\right] \end{aligned}

\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{-4}\left[\begin{array}{ccc} 1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 6 \end{array}\right] \\ &=\frac{-1}{4}\left[\begin{array}{c} 2+0-6 \\ 8-4+0 \\ -18+4+6 \end{array}\right] \quad \text { Mention the sign of matrix properly } \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{-1}{4}\left[\begin{array}{c} -4 \\ 4 \\ 8 \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]} \\ &x=1, y=-1, z=2 \end{aligned}