#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (vi)

Inconsistent

Given:

$x+y-2 z=5 \quad, \quad x-2 y+z=-2 \quad, \quad-2 x+y+z=4$

Hint:

Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.

Solution:

The given system of equations can be expressed as follows

\begin{aligned} &A X=B\\ &\text { Here, }\\ &A=\left[\begin{array}{ccc} 1 & 1 & -2 \\ 1 & -2 & 1 \\ -2 & 1 & 1 \end{array}\right] \quad, \quad X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{c} 5 \\ -2 \\ 4 \end{array}\right] \end{aligned}

\begin{aligned} &\text { Now, }\\ &\begin{aligned} |A|=\left|\begin{array}{ccc} 1 & 1 & -2 \\ 1 & -2 & 1 \\ -2 & 1 & 1 \end{array}\right| &=1(-2-1)-1(1+2)-2(1-4) \\ &=-3-3+6 \\ &=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 & 1 \\ 1 & 1 \end{array}\right|=-3 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 1 & -2 \\ 1 & 1 \end{array}\right|=-3 \\ &C_{12}=-1^{1+2}\left|\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right|=-3 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right|=-3 \end{aligned}

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

\begin{aligned} &\operatorname{adjA}=\left[\begin{array}{rrr} -3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3 \end{array}\right]^{T} \\ &(\operatorname{adj} A) B=\left[\begin{array}{rrr} -3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3 \end{array}\right]\left[\begin{array}{c} 5 \\ -2 \\ 4 \end{array}\right] \\ &=\left[\begin{array}{r} -15+6-12 \\ -15+6-12 \\ -15+6-12 \end{array}\right] \\ &=\left[\begin{array}{r} -21 \\ -21 \\ -21 \end{array}\right] \neq 0 \end{aligned}

Hence, the given system of equation is inconsistent.