UPSC Maths 2025 Paper 1 Q2a — Solution

15 marks · Section A

Question

Let $T : \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation such that $T(1, 1, -1) = (1, 0)$ , $T(4, 1, 1) = (0, 1)$ and $T(1, -1, 2) = (1, 1)$ . Find $T$ .

Technique

A linear map is determined by its action on a basis. Since the three given inputs form a basis, recover the standard matrix as $[T]=[\text{images}]\,[\text{basis}]^{-1}$ .

Solution

Step 1 — Verify the inputs form a basis.

Let $v_1=(1,1,-1)$ , $v_2=(4,1,1)$ , $v_3=(1,-1,2)$ . Put them as columns of $B=\begin{pmatrix}1 & 4 & 1\\ 1 & 1 & -1\\ -1 & 1 & 2\end{pmatrix},\qquad \det B = 1(2+1)-4(2-1)+1(1+1)=3-4+2=1\neq0.$ So $\{v_1,v_2,v_3\}$ is a basis of $\mathbb{R}^3$ and $T$ is uniquely determined.

Step 2 — Standard matrix of $T$ .

If $A=[T]$ is the $2\times3$ standard matrix, then $A v_i =$ (image $_i$ ). Stacking images as columns of $C=\begin{pmatrix}1 & 0 & 1\\0 & 1 & 1\end{pmatrix}$ , we have $A\,B = C$ , hence $A = C\,B^{-1}.$

Compute $B^{-1}$ (with $\det B=1$ ): $B^{-1}=\begin{pmatrix}3 & -7 & -5\\ -1 & 3 & 2\\ 2 & -5 & -3\end{pmatrix}.$

Then $A = \begin{pmatrix}1 & 0 & 1\\0 & 1 & 1\end{pmatrix}\begin{pmatrix}3 & -7 & -5\\ -1 & 3 & 2\\ 2 & -5 & -3\end{pmatrix} = \begin{pmatrix}5 & -12 & -8\\ 1 & -2 & -1\end{pmatrix}.$

Step 3 — Write $T$ explicitly.

$T(x,y,z) = (5x - 12y - 8z,\; x - 2y - z).$

Answer

$\boxed{\;T(x,y,z) = \big(\,5x - 12y - 8z,\;\; x - 2y - z\,\big).\;}$