← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q3a — Step-by-Step Solution

12 marks · Section A

Matrix of a linear transformation · Linear Algebra · asked 10× in 14 yrs · Read the full method →

Question

Let $V=\mathbb R^3$ and $T\in A(V)$, for all $a_i\in A(V)$, be defined by

What is the matrix $T$ relative to the basis $V_1=(1,0,1)$, $V_2=(-1,2,1)$, $V_3=(3,-1,1)$?

Technique

Standard change-of-basis: $[T]_B=P^{-1}[T]_{\text{std}}P$, computed concretely by applying $T$ to each $V_i$ then expressing in basis $B$.

Solution

Strategy. Compute $T(V_i)$ for $i=1,2,3$, then express each as $\sum_j c_{ji}V_j$. The matrix $[T]_B$ has columns equal to the coordinate column-vectors $(c_{1i},c_{2i},c_{3i})^T$.

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

$T(V_1)=T(1,0,1)=(2(1)+5(0)+1,\;-3(1)+0-1,\;-(1)+0+3)=(3,-4,2)$.

$T(V_2)=T(-1,2,1)=(-2+10+1,\;3+2-1,\;1+4+3)=(9,4,8)$.

$T(V_3)=T(3,-1,1)=(6-5+1,\;-9-1-1,\;-3-2+3)=(2,-11,-2)$.

Step 2 — Express each $T(V_i)$ in basis $\{V_1,V_2,V_3\}$

Solve $T(V_i)=c_1 V_1+c_2 V_2+c_3 V_3$.

The basis change matrix $P=[V_1\,V_2\,V_3]=\begin{pmatrix}1 & -1 & 3\\ 0 & 2 & -1\\ 1 & 1 & 1\end{pmatrix}$. We need $P^{-1}\,T(V_i)$ for each $i$.

Compute $P^{-1}$

$\det P$: expand along column 1.

$\det P=1\det\begin{pmatrix}2 & -1\\ 1 & 1\end{pmatrix}-0+1\det\begin{pmatrix}-1 & 3\\ 2 & -1\end{pmatrix}=1(2+1)+1(1-6)=3-5=-2$.

So $\det P=-2$.

Cofactor matrix (transpose for adjugate):

• $C_{11}=\det\begin{pmatrix}2 & -1\\ 1 & 1\end{pmatrix}=3$.
• $C_{12}=-\det\begin{pmatrix}0 & -1\\ 1 & 1\end{pmatrix}=-(0+1)=-1$.
• $C_{13}=\det\begin{pmatrix}0 & 2\\ 1 & 1\end{pmatrix}=-2$.
• $C_{21}=-\det\begin{pmatrix}-1 & 3\\ 1 & 1\end{pmatrix}=-(-1-3)=4$.
• $C_{22}=\det\begin{pmatrix}1 & 3\\ 1 & 1\end{pmatrix}=1-3=-2$.
• $C_{23}=-\det\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix}=-(1+1)=-2$.
• $C_{31}=\det\begin{pmatrix}-1 & 3\\ 2 & -1\end{pmatrix}=1-6=-5$.
• $C_{32}=-\det\begin{pmatrix}1 & 3\\ 0 & -1\end{pmatrix}=-(-1)=1$.
• $C_{33}=\det\begin{pmatrix}1 & -1\\ 0 & 2\end{pmatrix}=2$.

$\operatorname{adj}(P)=C^T=\begin{pmatrix}3 & 4 & -5\\ -1 & -2 & 1\\ -2 & -2 & 2\end{pmatrix}$.

$P^{-1}=\dfrac{1}{-2}\operatorname{adj}(P)=\dfrac{1}{-2}\begin{pmatrix}3 & 4 & -5\\ -1 & -2 & 1\\ -2 & -2 & 2\end{pmatrix}=\begin{pmatrix}-3/2 & -2 & 5/2\\ 1/2 & 1 & -1/2\\ 1 & 1 & -1\end{pmatrix}$.

Sanity check $PP^{-1}=I$

Row 1 of $P$ times col 1 of $P^{-1}$: $1(-3/2)+(-1)(1/2)+3(1)=-3/2-1/2+3=1$ ✓.

Row 1 of $P$ times col 2 of $P^{-1}$: $1(-2)+(-1)(1)+3(1)=-2-1+3=0$ ✓.

Row 1 of $P$ times col 3 of $P^{-1}$: $1(5/2)+(-1)(-1/2)+3(-1)=5/2+1/2-3=0$ ✓.

Step 3 — Apply $P^{-1}$ to each $T(V_i)$

For $T(V_1)=(3,-4,2)$:

For $T(V_2)=(9,4,8)$:

Wait, recompute: $-27/2-8+20=-27/2+12=(-27+24)/2=-3/2$.

So column 2 = $(-3/2,\;9/2,\;5)^T$. Let me double-check column 2:

$(-3/2)(9)=-27/2$. $(-2)(4)=-8$. $(5/2)(8)=20$. Sum: $-27/2-8+20=-27/2+12=-27/2+24/2=-3/2$ ✓.

$(1/2)(9)=9/2$. $(1)(4)=4$. $(-1/2)(8)=-4$. Sum: $9/2+4-4=9/2$ ✓.

$(1)(9)=9$. $(1)(4)=4$. $(-1)(8)=-8$. Sum: $9+4-8=5$ ✓.

So column 2 = $(-3/2,9/2,5)^T$.

For $T(V_3)=(2,-11,-2)$:

Step 4 — Assemble the matrix

Answer