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

\begin{aligned} x=\frac{5}{3}, y=\frac{3 k-4}{3} \text { and } z=k\\ \end{aligned}

Given:

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

\begin{aligned} &A X=B \\ &\qquad A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right], X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ &{\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]} \\ &\begin{aligned} |A| &=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right| \\ &=1(2-2)+1(4-1)+1(-4+1) \\ &=0+3-3=0 \\ |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}=(-1)^{1+1}\left|\begin{array}{cc} 1 & -1 \\ -2 & 2 \end{array}\right|=0 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right|=-3 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & 1 \\ -1 & -2 \end{array}\right|=-3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 1 \\ 2 & 2 \end{array}\right|=0 \end{aligned}

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

\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 0 & -3 & -3 \\ 0 & 3 & 3 \\ 0 & 3 & 3 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3 \end{array}\right] \end{aligned}

\begin{aligned} (\text { adj } A) B &=\left[\begin{array}{ccc} 0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ &=\left[\begin{array}{c} 0 \\ -9+6+3 \\ -9+6+3 \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=3-k \text { and } 2 x+y=2+k\\ &\left[\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3-k \\ 2+k \end{array}\right]\\ &\text { Now, }\\ &|A|=\left|\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right|\\ &=1+2\\ &=3 \neq 0 \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left|\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{3}\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right] \\ X=& A^{-1} B \end{aligned}

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

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{5}{3} \\ \frac{3 k-4}{3} \end{array}\right]} \\ &x=\frac{5}{3}, y=\frac{3 k-4}{3} \text { and } z=k \end{aligned}

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

\begin{aligned} Thus,\; x=\frac{5}{3}, y=\frac{3 k-4}{3} \text { and } z=k \end{aligned}

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