Given that the system of equations $:x + 4y + 8z = 16$---->(1)
$3x + 2y + 4z = 12$ ----->(2)
$4x + y + 2z = 10$-------->(3)
Now we can in the form of Matrix $A=\begin{bmatrix} 1 &4 &8 \\ 3 &2 &4 \\ 4 &1 &2 \end{bmatrix}$ and $B=\begin{bmatrix} 16\\12 \\ 10 \end{bmatrix}$
Let us write in the form of Augmented Matrix $[A:B]$=$\begin{bmatrix} 1 &4 &8:16 \\ 3 &2 &4:12 \\ 4 &1 &2:10 \end{bmatrix}$
Now perform some operation on $R_{3}$
$R_{3}\rightarrow R_{3}-R_{2}$
$[A:B]$=$\begin{bmatrix} 1 &4 &8:16 \\ 3 &2 &4:12 \\ 1 &-1 &-2:-2 \end{bmatrix}$
Now perform some operation on $R_{2}$
$R_{2}\rightarrow R_{2}-R_{1}$
$[A:B]$=$\begin{bmatrix} 1 &4 &8:16 \\ 2 &-2 &-4:-4 \\ 1 &-1 &-2:-2 \end{bmatrix}$
Now again perform some operation on $R_{2}$
$R_{2}\rightarrow \frac{1}{2}R_{2}$
$[A:B]$=$\begin{bmatrix} 1 &4 &8:16 \\ 1 &-1 &-2:-2 \\ 1 &-1 &-2:-2 \end{bmatrix}$
Now perform some operation on $R_{3}$
$R_{3}\rightarrow R_{3}-R_{2}$
$[A:B]$=$\begin{bmatrix} 1 &4 &8:16 \\ 1 &-1 &-2:-2 \\ 0 &0 &0: 0\end{bmatrix}$
Rank of $A=2$ and Rank of $[A: B]=2,$ which is equal.
$r(A)=r([A:B])=2$ and number of unknowns(Variables)$=3$
$r(A)=r([A:B]) <Unknowns$$(Variables)$ [This is the condition of infinitely many numbers of solutions]
$2<3$
Case$(1)$ If $r(A)\neq r([A:B])$ there is no solution.
Case$(2)$ If $r(A)= r([A:B])=Unknowns$$(No.$ $of$ $variables)$ there are Unique Solution.
Case$(3)$ If $r(A)=r([A:B]) <Unknowns$$(Variables)$ there are Infinitely many solutions.
In this case, if we want to know how to get many numbers of the solution:
Let say number of unknowns(Variables)$=n$ and rank of $r(A)=r([A:B])=r$,we can assign $n-r$ Linearly Independent value to the variables,and find out the many solutions.
