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

\begin{aligned} x=\frac{5-3 k}{2} \text { and } y=k\\ \end{aligned}

Given:

\begin{aligned} &2 x+3 y=5\\ &6 x+9 y=15 \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+3 y=5\; \; \; \; \; \; ......(i)\\ &6 x+9 y=15\; \; \; \; \; ......(ii) \end{aligned} \\ AX=B \\ Where

\begin{aligned} &A=\left[\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{c} 5 \\ 15 \end{array}\right] \\ &{\left[\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 5 \\ 15 \end{array}\right]} \\ &\begin{aligned} |A| &=\left|\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right| \\ &=18-18 \\ |A| &=0 \end{aligned} \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}=9, C_{12}=-6, C_{21}=-3, C_{22}=2 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 9 & -6 \\ -3 & 2 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 9 & -3 \\ -6 & 2 \end{array}\right] \end{aligned}

\begin{aligned} (\operatorname{adjA}) B &=\left[\begin{array}{cc} 9 & -3 \\ -6 & 2 \end{array}\right]\left[\begin{array}{c} 5 \\ 15 \end{array}\right] \\ &=\left[\begin{array}{c} 45-45 \\ -30+30 \end{array}\right] \\ &=\left[\begin{array}{l} 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), We get

\begin{aligned} &2 x+3 k=5 \\ &2 x=5-3 k \\ &x=\frac{5-3 k}{2} \\ &\text { And } y=k \end{aligned}

The values of x and y satisfy the third equation.

Thus\; \; \begin{aligned} x=\frac{5-3 k}{2} \text { and } y=k\\ \end{aligned}

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