Question-8

Let \(Ax=b\) be a system of linear equations.

\[ A=\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix},\,\,\,x=\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix},\,\,\,\,b=\begin{bmatrix}b_{1}\\ b_{2}\\ b_{3}\\ b_{4} \end{bmatrix} \]

First, note that \(A\) is in rref. We first identify the pivots and the pivot columns:

\[ \begin{bmatrix}\boldsymbol{1} & 0 & 0 & 0\\ 0 & 0 & \boldsymbol{1} & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \]

The first and third columns are pivot columns. Hence, \(x_{1},x_{3}\) are dependent and \(x_{2},x_{4}\) are independent variables. For the system to be consistent \(b_{3}=b_{4}=0\). Let us now work with this special case:

\[ A=\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix},\,\,\,x=\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix},\,\,\,\,b=\begin{bmatrix}b_{1}\\ b_{2}\\ 0\\ 0 \end{bmatrix} \]

We can give arbitrary values to \(x_{2}\) and \(x_{4}\) and then solve for \(x_{1}\) and \(x_{3}\):

\[ \begin{aligned} x_{2} & =t_{2}\\ x_{4} & =t_{4}\\ x_{3} & =b_{2}-t_{4}\\ x_{1} & =b_{1} \end{aligned} \]

The set of all solutions to the system is given by:

\[ S=\left\{ \left(b_{1},t_{2},b_{2}-t_{4},t_{4}\right):t_{2},t_{4}\in\mathbb{R}\right\} \]