UPSC 2024 Maths Optional Paper 1 Q2a — Step-by-Step Solution

15 marks · Section A

Inverse of a matrix (adjoint and row reduction) · Linear Algebra · asked 3× in 14 yrs · Read the full method →

Question

Consider a linear operator $T$ on $\mathbb{R}^3$ over $\mathbb{R}$ defined by $T(x,y,z)=(2x,\,4x-y,\,2x+3y-z)$ . Is $T$ invertible? If yes, justify your answer and find $T^{-1}$ .

Technique

Read off the triangular matrix of $T$ ; determinant is the product of diagonal entries; invert by back-substitution.

Solution

Step 1 — Matrix of $T$ .

[T]=\begin{pmatrix}2 & 0 & 0\\ 4 & -1 & 0\\ 2 & 3 & -1\end{pmatrix}.

Step 2 — Determinant.

Lower-triangular: $\det[T]=2\cdot(-1)\cdot(-1)=2\ne 0$ . Therefore $T$ is invertible.

Step 3 — Find $T^{-1}$ .

Solve $T(x,y,z)=(a,b,c)$ :

$2x=a\;\Rightarrow\; x=a/2$ .
$4x-y=b\;\Rightarrow\; y=4x-b=2a-b$ .
$2x+3y-z=c\;\Rightarrow\; z=2x+3y-c=a+3(2a-b)-c=7a-3b-c$ .

Answer

\boxed{\;T^{-1}(a,b,c)=\Bigl(\tfrac{a}{2},\;2a-b,\;7a-3b-c\Bigr).\;}