UPSC 2023 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Rank and nullity; rank-nullity theorem · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Find the rank and nullity of the linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$ given by $T(x,y,z)=(x+z,\,x+y+2z,\,2x+y+3z)$ .

Standard matrix-of- $T$ + row reduction. Rank from echelon form, nullity from rank–nullity.

Step 1 — Matrix of $T$ in standard basis.

[T]=\begin{bmatrix}1 & 0 & 1\\1 & 1 & 2\\2 & 1 & 3\end{bmatrix}.

Step 2 — Row-reduce.

$R_2\to R_2-R_1$ : $(0,1,1)$ . $R_3\to R_3-2R_1$ : $(0,1,1)$ . $R_3\to R_3-R_2$ : $(0,0,0)$ .

\begin{bmatrix}1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 0\end{bmatrix}.

Two non-zero rows, so $\operatorname{rank}T=2$ .

Step 3 — Nullity.

By the rank–nullity theorem on $T:\mathbb R^3\to\mathbb R^3$ :

\operatorname{nullity}T=3-\operatorname{rank}T=1.

\boxed{\;\operatorname{rank}T=2,\quad\operatorname{nullity}T=1.\;}