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UPSC 2021 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Matrix of a linear transformation · Linear Algebra · asked 10× in 13 yrs · Read the full method →

Question

Find the matrix associated with the linear operator $T(a,b,c)=(a+b,a-b,2c)$ on $V_3(\mathbb R)$ with respect to ordered basis $B=\{(0,1,1),(1,0,1),(1,1,0)\}$.

Technique

Apply $T$ to each basis vector; solve linear system to express in basis $B$; columns of $[T]_B$ are coordinate vectors.

Solution

Strategy. Compute $T(V_i)$ for each basis vector; express each in basis $B$; columns of $[T]_B$ are the coordinate vectors.

Step 1 — Compute $T(V_i)$ in standard coordinates

$T(V_1)=T(0,1,1)=(0+1,\;0-1,\;2)=(1,-1,2)$.

$T(V_2)=T(1,0,1)=(1+0,\;1-0,\;2)=(1,1,2)$.

$T(V_3)=T(1,1,0)=(1+1,\;1-1,\;0)=(2,0,0)$.

Step 2 — Express each $T(V_i)$ as $\sum c_j V_j$

Solve $T(V_1)=c_1 V_1+c_2 V_2+c_3 V_3=(c_2+c_3,\;c_1+c_3,\;c_1+c_2)$.

System:

• $c_2+c_3=1$
• $c_1+c_3=-1$
• $c_1+c_2=2$

Add all: $2(c_1+c_2+c_3)=2\Rightarrow c_1+c_2+c_3=1$.

Subtract first: $c_1=1-1=0$. Subtract second: $c_2=1-(-1)=2$. Subtract third: $c_3=1-2=-1$.

Check: $c_2+c_3=2-1=1$ ✓; $c_1+c_3=0-1=-1$ ✓; $c_1+c_2=0+2=2$ ✓.

Column 1 of $[T]_B$: $(0, 2, -1)^T$.

For $T(V_2)=(1,1,2)$:

• $c_2+c_3=1$
• $c_1+c_3=1$
• $c_1+c_2=2$

Sum: $2(c_1+c_2+c_3)=4\Rightarrow c_1+c_2+c_3=2$.

$c_1=2-1=1$, $c_2=2-1=1$, $c_3=2-2=0$.

Column 2: $(1, 1, 0)^T$.

For $T(V_3)=(2,0,0)$:

• $c_2+c_3=2$
• $c_1+c_3=0$
• $c_1+c_2=0$

Sum: $2(c_1+c_2+c_3)=2\Rightarrow c_1+c_2+c_3=1$.

$c_1=1-2=-1$, $c_2=1-0=1$, $c_3=1-0=1$.

Check: $c_2+c_3=1+1=2$ ✓; $c_1+c_3=-1+1=0$ ✓; $c_1+c_2=-1+1=0$ ✓.

Column 3: $(-1, 1, 1)^T$.

Step 3 — Matrix

Answer