If the system of linear equations : … is consistent and has infinite number of solutions, then :

Given system of equations can be written in matrix form as $\mathrm{AX}=\mathrm{B}$AX=B where $\mathrm{A}=\left(\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right)$A=​112​235​35a​​ and $\mathrm{B}=\left(\begin{array}{l}6 \\ 9 \\ b\end{array}\right)$B=​69b​​ Since, system is consistent and has infinitely many solutions $\therefore(\operatorname{adj} . \mathrm{A}) \mathrm{B}=0$∴(adj.A)B=0 $$\begin{aligned} & \Rightarrow\left(\begin{array}{ccc} 3 a-25 & 15-2 a & 1 \\ 10-a & a-6 & -2 \\ -1 & -1 & 1 \end{array}\right)\left(\begin{array}{l} 6 \\ 9 \\ b \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right) \\ & \Rightarrow-6-9+b=0 \Rightarrow b=15 \\ & \text { and } 6(10-a)+9(a-6)-2(b)=0 \\ & \Rightarrow 60-6 a+9 a-54-30=0 \\ & \Rightarrow 3 a=24 \Rightarrow a=8 \end{aligned}$$​⇒​3a−2510−a−1​15−2aa−6−1​1−21​​​69b​​=​000​​⇒−6−9+b=0⇒b=15 and 6(10−a)+9(a−6)−2(b)=0⇒60−6a+9a−54−30=0⇒3a=24⇒a=8​ Hence, $a=8, b=15$a=8,b=15.