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

$x=\frac{7-16 k}{11}, y=\frac{3+k}{11} \text { and } z=k$

Given:

\begin{aligned} &5 x+3 y+7 z=4 \\ &3 x+26 y+2 z=9 \\ &7 x+2 y+10 z=5 \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} 5 x+3 y+7 z=4 \; \; \; \; \; ......(i)\\ 3 x+26 y+2 z=9 \; \; \; \; \; ......(ii)\\ 7 x+2 y+10 z=5 \; \; \; \; \; ......(iii) \end{aligned} \\ \\ AX=B \\ \\ Where

\begin{aligned} &A=\left[\begin{array}{ccc} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 4 \\ 9 \\ 5 \end{array}\right] \\ &{\left[\begin{array}{ccc} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4 \\ 9 \\ 5 \end{array}\right]} \end{aligned}

\begin{aligned} |A| &=\left|\begin{array}{ccc} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{array}\right| \\ &=5(260-4)-3(30-14)+7(6-182) \\ &=1280-48-1232=0 \\ |A| &=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} 26 & 2 \\ 2 & 10 \end{array}\right|=256 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & 2 \\ 7 & 10 \end{array}\right|=-16 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 26 \\ 7 & 2 \end{array}\right|=-176 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 3 & 7 \\ 2 & 10 \end{array}\right|=-16 \end{aligned}

\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 5 & 7 \\ 7 & 10 \end{array}\right|=1 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 5 & 3 \\ 7 & 2 \end{array}\right|=11 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 3 & 7 \\ 26 & 2 \end{array}\right|=-176 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 5 & 7 \\ 3 & 2 \end{array}\right|=11 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 5 & 3 \\ 3 & 26 \end{array}\right|=121 \end{aligned}

\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121 \end{array}\right] \end{aligned}

\begin{aligned} (\operatorname{adj} A) B &=\left[\begin{array}{ccc} 256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121 \end{array}\right]\left[\begin{array}{l} 4 \\ 9 \\ 5 \end{array}\right] \\ &=\left[\begin{array}{c} 1024-144-880 \\ -64+9+55 \\ -704+9+605 \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} &5 x+3 y=4-7 k \text { and } 3 x+26 y=9-2 k \\ &{\left[\begin{array}{cc} 5 & 3 \\ 3 & 26 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4-7 k \\ 9-2 k \end{array}\right]} \end{aligned}

\begin{aligned} |A| &=\left|\begin{array}{cc} 5 & 3 \\ 3 & 26 \end{array}\right| \\ &=130-9 \\ &=121 \neq 0 \\ \operatorname{adj} A &=\left|\begin{array}{cc} 26 & -3 \\ -3 & 5 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{121}\left[\begin{array}{cc} 26 & -3 \\ -3 & 5 \end{array}\right] \end{aligned}

\begin{aligned} &X=A^{-1} B \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{121}\left[\begin{array}{cc} 26 & -3 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 4-7 k \\ 9-2 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{121}\left[\begin{array}{l} 104-182 k-27+6 k \\ -12+21 k+45-10 k \end{array}\right]} \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{77-176 k}{121} \\ \frac{33+11 k}{121} \end{array}\right]} \\ &x=\frac{11(7-16 k)}{121}, y=\frac{11(3+k)}{121} \text { and } z=k \\ &x=\frac{7-16 k}{11}, y=\frac{3+k}{11} \text { and } z=k \end{aligned}

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

$Thus\; x=\frac{7-16 k}{11}, y=\frac{3+k}{11} \text { and } z=k$

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