UPSC 2023 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Linear dependence and independence · Linear Algebra · asked 3× in 14 yrs · Read the full method →

Question

Let $V_1=(2,-1,3,2)$ , $V_2=(-1,1,1,-3)$ and $V_3=(1,1,9,-5)$ be three vectors of the space $\mathbb{R}^4$ . Does $(3,-1,0,-1)\in\operatorname{span}\{V_1,V_2,V_3\}$ ? Justify your answer.

Technique

Reduce the spanning system to its first three equations; solve two for $a,b$ in terms of $c$ ; substitute into the third; check consistency.

Solution

Step 1 — Set up the linear system.

We seek scalars $a,b,c$ with $aV_1+bV_2+cV_3=(3,-1,0,-1)$ , i.e.,

\begin{aligned}2a-b+c&=3\\-a+b+c&=-1\\3a+b+9c&=0\\2a-3b-5c&=-1.\end{aligned}

Step 2 — Reduce.

Add the first two equations: $a+2c=2$ , so $a=2-2c$ .

Substitute into the second: $-(2-2c)+b+c=-1\Rightarrow b=-1+2-2c-c=1-3c$ .

Substitute $a,b$ into the third equation:

3(2-2c)+(1-3c)+9c=6-6c+1-3c+9c=7.

For consistency we’d need $7=0$ . Contradiction.

Step 3 — Conclude.

The system has no solution, so

Answer

\boxed{\;(3,-1,0,-1)\notin\operatorname{span}\{V_1,V_2,V_3\}.\;}