If the system of simultaneous linear equations $x+y+z=a, x-y+b z=2$, $2 x+3 y-z=1$ has infinitely many solutions, then $b-5 a=$

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Given that system of linear equations $\begin{aligned} x+y+z & =a \quad \ldots (i)\ x-y+b z & =2 \quad \ldots (ii) \ 2 x+3 y-z & =1 \quad \ldots (iii) \end{aligned}$ has infinitely many solutions, $\begin{aligned} & \therefore \quad\left[\begin{array}{ccc} 1 & 1 & 1 \ 1 & -1 & b \ 2 & 3 & -1 \end{array}\right]=0 \ & \Rightarrow \quad 1(1-3 b)-1(-1-2 b)+1(3+2=0 \ & \Rightarrow \quad 1-3 b+1+2 b+5=0 \ & \Rightarrow \ & -b=-7 \ & \Rightarrow \ & b=7 \end{aligned}$ Adding Eqs. (i) and (ii), we get $\quad \begin{aligned} 2 x+8 z & =a+2 \ \Rightarrow x+4 z & =\frac{a+2}{2} \quad \ldots (iv) \end{aligned}$ Multiply Eq. (i) by 3 then subtract Eq. (iii) from it, $\begin{aligned} 3 x+3 y+3 z & =3 a \ -2 x \pm 3 y \mp z & =-1 \ \hline x+4 z & =3 a-1 \quad \ldots (v) \end{aligned}$ By Eqs. (iv) and (v), we get $\begin{array}{rlrl} & & \frac{a+2}{2} & =3 a-1 \ \Rightarrow & & a+2 & =6 a-2 \ \Rightarrow & 6 a-a & =4 \ \Rightarrow & & 5 a & =4 \ \therefore & b-5 a & =7-4=3 \end{array}$

If the system of simultaneous linear equations \(x+y+z=a, x-y+b z=2\), \(2 x+3 y-z=1\) has infinitely many…

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