#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (vi)

\begin{aligned} x=\frac{1-2 k}{2}, y=k \text { and } z=0 \end{aligned}

Given:

\begin{aligned} &2 x+2 y-2 z=1\\ &4 x+4 y-z=2\\ &6 x+6 y+2 z=3 \end{aligned}

Hint:

Consistent equation means two or more equations that are possible to solve based on using set of values for the variables.

Solution: Here,

\begin{aligned} &2 x+2 y-2 z=1\; \; \; \; \; .....(i)\\ &4 x+4 y-z=2\; \; \; \; \; .....(ii)\\ &6 x+6 y+2 z=3\; \; \; \; \; .....(iii) \end{aligned}

\begin{aligned} &A X=B\\ &\text { Where }\\ &A=\left[\begin{array}{ccc} 2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{ccc} 2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ &|A|=\left|\begin{array}{ccc} 2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2 \end{array}\right| \\ &= 2(8+6)-2(8+6)-2(24-24) \\ & =28-28=0 \end{aligned}

So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because

$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0$

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

\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 4 & -1 \\ 6 & 2 \end{array}\right|=14 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & -1 \\ 6 & 2 \end{array}\right|=-14 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 4 & 4 \\ 6 & 6 \end{array}\right|=0 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 2 & -2 \\ 6 & 2 \end{array}\right|=-16 \end{aligned}

\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & -2 \\ 6 & 2 \end{array}\right|=16 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 2 & 2 \\ 6 & 6 \end{array}\right|=0 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right|=6 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right|=-6 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 2 & 2 \\ 4 & 4 \end{array}\right|=0 \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} -14 & -14 & 0 \\ -16 & 16 & 0 \\ 6 & -6 & 0 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 14 & -16 & 6 \\ -14 & 16 & -6 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}

\begin{aligned} (\text { adjA }) B &=\left[\begin{array}{ccc} 14 & -16 & 6 \\ -14 & 16 & -6 \\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] \\ &=\left[\begin{array}{c} 14-32+18 \\ -14+32-18 \\ 0 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{aligned}

If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.

Thus, AX = B has infinitely many solutions

Substituting y = k in eqn (i) and eqn (ii), We get

\begin{aligned} &2 x+2 z=1-2 k \text { and } 4 x+z=2-4 k \\ &{\left[\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 1-2 k \\ 2-4 k \end{array}\right]} \\ &\begin{aligned} |A| &=\left|\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right| \\ &=-2+8 \\ &=6 \neq 0 \end{aligned} \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left|\begin{array}{rr} -1 & 2 \\ -4 & 2 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{6}\left[\begin{array}{rr} -1 & 2 \\ -4 & 2 \end{array}\right] \\ X=& A^{-1} B \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{6}\left[\begin{array}{ll} -1 & 2 \\ -4 & 2 \end{array}\right]\left[\begin{array}{l} 1-2 k \\ 2-4 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{6}\left[\begin{array}{l} -1+2 k+4-8 k \\ -4+8 k+4-8 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{3-6 k}{6} \\ 0 \end{array}\right]} \end{aligned}

$x=\frac{1}{2} -k ,\; y=k\; and\; z=0$

The values of x and y and z satisfy the third equation.

$x=\frac{1}{2} -k ,\; y=k\; and\; z=0$

Where ‘ $k$ ’ is a real number satisfy the given system of equations.