If the system of linear equations : $$ \begin{aligned} & x_1+2 x_2+3 x_3=6 \\ & x_1+3 x_2+5 x_3=9 \\ & 2…
If the system of linear equations :
$$
\begin{aligned}
& x_1+2 x_2+3 x_3=6 \\
& x_1+3 x_2+5 x_3=9 \\
& 2 x_1+5 x_2+a x_3=b
\end{aligned}
$$
is consistent and has infinite number of solutions, then :
$a=8, b$ can be any real number
$b=15, a$ can be any real number
$a \in R-\{8\}$ and $b \in R-\{15\}$
$a=8, b=15$
Solution
Given system of equations can be written in matrix form as $\mathrm{AX}=\mathrm{B}$ where $\mathrm{A}=\left(\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{l}6 \\ 9 \\ b\end{array}\right)$
Since, system is consistent and has infinitely many solutions
$\therefore(\operatorname{adj} . \mathrm{A}) \mathrm{B}=0$
$
\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}
$
Hence, $a=8, b=15$