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UPSC 2015 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Linear dependence and independence · Linear Algebra · asked 3× in 13 yrs · Read the full method →

Question

The vectors $V_1=(1,1,2,4)$, $V_2=(2,-1,-5,2)$, $V_3=(1,-1,-4,0)$ and $V_4=(2,1,1,6)$ are linearly independent. Is it true? Justify your answer.

Technique

Row reduction of the matrix of row vectors; identify rank and exhibit a kernel vector.

Solution

Strategy. Form the $4\times 4$ matrix whose rows are $V_1,V_2,V_3,V_4$. The four vectors are linearly independent iff this matrix has full rank 4 iff $\det\ne 0$. We compute the determinant by row reduction.

Step 1 — Set up the matrix

Step 2 — Row reduce

$R_2\to R_2-2R_1,\;R_3\to R_3-R_1,\;R_4\to R_4-2R_1$:

Factor $-3,-2,-1$ from $R_2,R_3,R_4$ respectively — but easier: note $R_2=3R_4$ and $R_3=2R_4$ (since $-3:-2:-1=3:2:1$, and $-9:-6:-3=3:2:1$, and $-6:-4:-2=3:2:1$). So $R_2,R_3,R_4$ are all scalar multiples of $(0,-1,-3,-2)$ — three linearly dependent rows.

Explicitly, $R_2\to R_2-3R_4=(0,0,0,0)$ and $R_3\to R_3-2R_4=(0,0,0,0)$, leaving

Step 3 — Read off rank

Rank $= 2$. Hence $\det M = 0$, and the four vectors are linearly dependent.

Step 4 — Exhibit an explicit linear relation

From the row reduction, $V_2-3V_4'=0$ where $V_4'$ is $V_4$ after the first elimination — easier to find directly. Solve $\alpha V_1+\beta V_2+\gamma V_3+\delta V_4=0$:

From the four coordinate equations,

Try $\alpha=0$: then $2\beta+\gamma+2\delta=0$, $-\beta-\gamma+\delta=0$, $2\beta+6\delta=0\Rightarrow\beta=-3\delta$. Substitute into first: $-6\delta+\gamma+2\delta=0\Rightarrow\gamma=4\delta$. Check second: $3\delta-4\delta+\delta=0$ ✓. Check third: $15\delta-16\delta+\delta=0$ ✓.

Take $\delta=1$: $(\alpha,\beta,\gamma,\delta)=(0,-3,4,1)$. So

Answer