UPSC Maths 2025 Paper 2 Q1e — Solution

10 marks · Section A

Question

How many basic solutions are there for the following system of equations? $2x_1 - x_2 + 3x_3 + x_4 = 6$ $4x_1 - 2x_2 - x_3 + 2x_4 = 10$ Find all of them. Furthermore, find the number of basic solutions, which are feasible/non-feasible/non-degenerate.

Technique

A basic solution chooses $m=2$ basic variables (an invertible $m\times m$ submatrix of columns), sets the other variables to zero, and solves. Count the nonsingular column-pairs out of $\binom{4}{2}=6$ , solve each, then classify.

Solution

The system has $m=2$ equations and $n=4$ unknowns. The coefficient matrix is $A = \begin{pmatrix} 2 & -1 & 3 & 1 \\ 4 & -2 & -1 & 2 \end{pmatrix},\qquad b = \begin{pmatrix}6\\10\end{pmatrix}.$

A basic solution is obtained by selecting $m=2$ columns forming a nonsingular $2\times2$ matrix $B$ (the basis), setting the other two variables to $0$ , and solving $B x_B = b$ . The maximum possible number is $\binom{4}{2}=6$ .

Step 1 — Identify which column pairs are nonsingular.

The four columns are $c_1=\begin{pmatrix}2\\4\end{pmatrix},\quad c_2=\begin{pmatrix}-1\\-2\end{pmatrix},\quad c_3=\begin{pmatrix}3\\-1\end{pmatrix},\quad c_4=\begin{pmatrix}1\\2\end{pmatrix}.$

The columns $c_1,c_2,c_4$ are all scalar multiples of $(1,2)^T$ : $c_1=2(1,2)^T$ , $c_2=-(1,2)^T$ , $c_4=(1,2)^T$ . Hence any pair drawn entirely from $\{c_1,c_2,c_4\}$ is linearly dependent (determinant $0$ ):

$(c_1,c_2)$ : $\det=0$ ,
$(c_1,c_4)$ : $\det=0$ ,
$(c_2,c_4)$ : $\det=0$ .

The remaining three pairs all contain $c_3$ , which is not parallel to $(1,2)^T$ , so each is nonsingular:

$(c_1,c_3)$ : $\det = 2(-1)-3(4) = -14 \neq 0$ ,
$(c_2,c_3)$ : $\det = (-1)(-1)-3(-2) = 7 \neq 0$ ,
$(c_3,c_4)$ : $\det = 3(2)-1(-1) = 7 \neq 0$ .

So exactly 3 basic solutions exist.

Step 2 — Compute the 3 basic solutions.

Basis $\{x_1,x_3\}$ ( $x_2=x_4=0$ ): $2x_1+3x_3=6,\quad 4x_1-x_3=10 \;\Rightarrow\; x_1=\tfrac{18}{7},\ x_3=\tfrac{2}{7}.$ $\mathbf{x}^{(1)}=\left(\tfrac{18}{7},\,0,\,\tfrac{2}{7},\,0\right).$

Basis $\{x_2,x_3\}$ ( $x_1=x_4=0$ ): $-x_2+3x_3=6,\quad -2x_2-x_3=10 \;\Rightarrow\; x_2=-\tfrac{36}{7},\ x_3=\tfrac{2}{7}.$ $\mathbf{x}^{(2)}=\left(0,\,-\tfrac{36}{7},\,\tfrac{2}{7},\,0\right).$

Basis $\{x_3,x_4\}$ ( $x_1=x_2=0$ ): $3x_3+x_4=6,\quad -x_3+2x_4=10 \;\Rightarrow\; x_3=\tfrac{2}{7},\ x_4=\tfrac{36}{7}.$ $\mathbf{x}^{(3)}=\left(0,\,0,\,\tfrac{2}{7},\,\tfrac{36}{7}\right).$

Step 3 — Classify.

Feasible (all components $\ge 0$ ): $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(3)}$ are feasible. $\mathbf{x}^{(2)}$ has $x_2=-36/7<0$ , so it is non-feasible.
Non-degenerate (all $m=2$ basic variables nonzero): in every one of the three solutions both basic variables are nonzero. So all $3$ are non-degenerate.

Answer

Number of basic solutions: 3 (out of a possible $\binom{4}{2}=6$ ; the other 3 column-pairs are singular because columns $c_1,c_2,c_4$ are mutually parallel).
The three basic solutions are $\left(\tfrac{18}{7},0,\tfrac{2}{7},0\right),\quad \left(0,-\tfrac{36}{7},\tfrac{2}{7},0\right),\quad \left(0,0,\tfrac{2}{7},\tfrac{36}{7}\right).$
Feasible: 2 $\left(\mathbf{x}^{(1)},\mathbf{x}^{(3)}\right)$ ; non-feasible: 1 $\left(\mathbf{x}^{(2)}\right)$ ; non-degenerate: 3 (degenerate: 0).