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#### Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (v) maths textbook solution

$x=k-2,\; y=8-2k\; and\; z=k$

Given:

\begin{aligned} &x+y+z=6 \\ &x+2 y+3 z=14 \\ &x+4 y+7 z=30 \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} &x+y+z=6 \; \; \; \; \; .....(i)\\ &x+2 y+3 z=14 \; \; \; \; \; .....(ii) \\ &x+4 y+7 z=30 \; \; \; \; \; .....(iii) \end{aligned}

\begin{aligned} &A X=B\\ &\text { Where }\\ &A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right] \end{aligned}

\begin{aligned} {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right]} \\ |A|=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right| \end{aligned}\\ =1(14-12)-1(7-3)+1(4-2) \\ =2-4+2=0 \\ \left | A \right |=0

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}{ll} 2 & 3 \\ 4 & 7 \end{array}\right|=2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 1 & 3 \\ 1 & 7 \end{array}\right|=-4 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 1 & 2 \\ 1 & 4 \end{array}\right|=2 \quad , \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 7 \end{array}\right|=-3 \end{aligned}

\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 7 \end{array}\right|=6 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 1 & 1 \\ 1 & 4 \end{array}\right|=-3 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right|=1 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 3 \end{array}\right|=-2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right|=1 \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 2 & -4 & 2 \\ -3 & 6 & -3 \\ 1 & -2 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1 \end{array}\right] \end{aligned}

\begin{aligned} (\operatorname{adj} A) B &=\left[\begin{array}{ccc} 2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right] \\ &=\left[\begin{array}{c} 12-42+30 \\ -24+84-60 \\ 12-42+30 \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 z = k in eqn (i) and eqn (ii), We get

\begin{aligned} &x+y=6-k \text { and } x+2 y=14-3 k\\ &\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 6-k \\ 14-3 k \end{array}\right]\\ &\text { Now, }\\ &|A|=\left|\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right|\\ &=2-1\\ &=1 \neq 0 \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left|\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{1}\left[\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right] \\ X=& A^{-1} B \end{aligned}

\begin{aligned} &{\left[\begin{array}{c} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{c} 6-k \\ 14-3 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{c} 12-2 k-14+3 k \\ -6+k+14-3 k \end{array}\right]} \\ &{\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{k-2}{1} \\ \frac{8-2 k}{1} \end{array}\right]} \end{aligned}

$x=k-2,\; y=8-2k\; and\; z=k$

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

$Thus\; x=k-2,\; y=8-2k\; and\; z=k$

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