Consider a consistent system of three linear equations in three variables. Discu

Step 1
Let the consistent system of three linear equations in three variables be given by
${\displaystyle a_{11}{x}_1+a_{12}{x}_2+a_{13}{x}_3=b_1}$
${\displaystyle a_{21}{x}_1+a_{22}{x}_1+a_{23}{x}_3=b_2}$
${\displaystyle a_{31}{x}_1+a_{32}{x}_1+a_{33}{x}_3=b_3}$
A) If thereduced form of the augmented coefficient matrix of the above consistent linear system has One leftmost 1 then that 1 must occur at the first row. Otherwise you can apply suitable row exchange to get that 1 at first row. Note that except the first row all other rows are zero because if a row is nonzero then we can divide that row by its first nonzero element to get a leftmost 1. Therefore in this case we may have following possibilities.
$\left[\begin{array}{ccc}1& \ast & \ast |\ast \\ 0& 0& 0|0\\ 0& 0& 0|0\end{array}\right]$
$\left[\begin{array}{ccc}0& 1& \ast |\ast \\ 0& 0& 0|0\\ 0& 0& 0|0\end{array}\right]$
$\left[\begin{array}{ccc}0& 0& 1|\ast \\ 0& 0& 0|0\\ 0& 0& 0|0\end{array}\right]$
Thus in each of the following cases the number of leftmost 1's in a reduced augmented coefficient matrix is two less than the number of variables in the system and therefore the system is dependent with two parameters and has infinitely many solutions.
Step 2
B) If the reduced form of the augmented coefficient matrix of the above consistent linear system has Two leftmost 1's then third row of the reduced form is always zero. Therefore we may have following possibilities.

$\left[\begin{array}{ccc}1& 0& \ast |\ast \\ 0& 1& \ast |\ast \\ 0& 0& 0|0\end{array}\right]$
$\left[\begin{array}{ccc}1& \ast & 0|\ast \\ 0& 0& 1|\ast \\ 0& 0& 0|0\end{array}\right]$
$\left[\begin{array}{ccc}0& 1& 0|\ast \\ 0& 0& 1|\ast \\ 0& 0& 0|0\end{array}\right]$
Since the number of leftmost 1's in a reduced augmented coefficient matrix is once less than the number of variables in the system and therefore the system is dependent with one parameters and has infinitely many solutions.
Step 3
If the reduced form of the augmented coefficient matrix of the above consistent linear system has Three leftmost 1's then it can be written as
$\left[\begin{array}{ccc}1& 0& 0|\ast \\ 0& 1& 0|\ast \\ 0& 0& 0|\ast \end{array}\right]$
Since the reduced form of the augmented coefficient matrix has three left most 1's which is equal to the number of variables of the lineart system, the linear system of equations is independent and hence has a unique solution.
D) Since we have a consistent system of three linear equations in three variables so the augmented matrix has three rows. In order to get four left most 1's we need at least four rows in the augmented matrix, which is impossible. Therefore, it is impossible to get a reduced form of augmented coefficient matrix with Four leftmost 1's.