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UPSC 2017 Maths Optional Paper 1 Q3a — Step-by-Step Solution 15 marks · Section A
Rank and nullity; rank-nullity theorem · Linear Algebra · asked 7× in 13 yrs · Read the full method →
Question
Consider the matrix mapping A : R 4 → R 3 A:\mathbb R^4\to\mathbb R^3 A : R 4 → R 3 A = ( 1 2 3 1 1 3 5 − 2 3 8 13 − 3 ) A=\begin{pmatrix}1 & 2 & 3 & 1\\ 1 & 3 & 5 & -2\\ 3 & 8 & 13 & -3\end{pmatrix} A = 1 1 3 2 3 8 3 5 13 1 − 2 − 3 A A A A A A
Technique
Row-reduce to rref; image basis = = =
Solution
Step 1 — Row-reduce A A A
A = ( 1 2 3 1 1 3 5 − 2 3 8 13 − 3 ) → R 3 → R 3 − 3 R 1 R 2 → R 2 − R 1 ( 1 2 3 1 0 1 2 − 3 0 2 4 − 6 ) → R 3 → R 3 − 2 R 2 ( 1 2 3 1 0 1 2 − 3 0 0 0 0 ) . A=\begin{pmatrix}1&2&3&1\\1&3&5&-2\\3&8&13&-3\end{pmatrix}\xrightarrow[R_3\to R_3-3R_1]{R_2\to R_2-R_1}\begin{pmatrix}1&2&3&1\\0&1&2&-3\\0&2&4&-6\end{pmatrix}\xrightarrow{R_3\to R_3-2R_2}\begin{pmatrix}1&2&3&1\\0&1&2&-3\\0&0&0&0\end{pmatrix}. A = 1 1 3 2 3 8 3 5 13 1 − 2 − 3 R 2 → R 2 − R 1 R 3 → R 3 − 3 R 1 1 0 0 2 1 2 3 2 4 1 − 3 − 6 R 3 → R 3 − 2 R 2 1 0 0 2 1 0 3 2 0 1 − 3 0 . → R 1 → R 1 − 2 R 2 ( 1 0 − 1 7 0 1 2 − 3 0 0 0 0 ) ( rref ) . \xrightarrow{R_1\to R_1-2R_2}\ \begin{pmatrix}1&0&-1&7\\0&1&2&-3\\0&0&0&0\end{pmatrix}\ (\text{rref}). R 1 → R 1 − 2 R 2 1 0 0 0 1 0 − 1 2 0 7 − 3 0 ( rref ) . Pivots are in columns 1 1 1 2 2 2 rank A = 2 \operatorname{rank}A=2 rank A = 2
Step 2 — Image (column space): basis and dimension
The image is spanned by the original columns corresponding to the pivot positions (columns 1 , 2 1,2 1 , 2
Image ( A ) = span { ( 1 1 3 ) , ( 2 3 8 ) } , dim = 2. \text{Image}(A)=\operatorname{span}\left\{\begin{pmatrix}1\\1\\3\end{pmatrix},\ \begin{pmatrix}2\\3\\8\end{pmatrix}\right\},\qquad \dim=2. Image ( A ) = span ⎩ ⎨ ⎧ 1 1 3 , 2 3 8 ⎭ ⎬ ⎫ , dim = 2. (These two are independent and every other column is their combination: col3 = _3= 3 = 1 + 2 _1+2\, 1 + 2 2 _2 2 1 ⋅ 1 + 2 ⋅ 2 = 5 ≠ 3 1\cdot1+2\cdot2=5\ne3 1 ⋅ 1 + 2 ⋅ 2 = 5 = 3 3 = − _3=- 3 = − 1 + 2 _1+2\, 1 + 2 2 _2 2 − 1 + 4 = 3 , − 1 + 6 = 5 , − 3 + 16 = 13 -1+4=3,\ -1+6=5,\ -3+16=13 − 1 + 4 = 3 , − 1 + 6 = 5 , − 3 + 16 = 13 4 = 7 _4=7\, 4 = 7 1 − 3 _1-3\, 1 − 3 2 _2 2
Step 3 — Kernel (null space): basis and dimension
By rank–nullity, dim ker A = 4 − rank A = 4 − 2 = 2 \dim\ker A=4-\operatorname{rank}A=4-2=2 dim ker A = 4 − rank A = 4 − 2 = 2 x 3 = s , x 4 = t x_3=s,\ x_4=t x 3 = s , x 4 = t
x 1 − x 3 + 7 x 4 = 0 ⇒ x 1 = s − 7 t , x 2 + 2 x 3 − 3 x 4 = 0 ⇒ x 2 = − 2 s + 3 t . x_1-x_3+7x_4=0\Rightarrow x_1=s-7t,\qquad x_2+2x_3-3x_4=0\Rightarrow x_2=-2s+3t. x 1 − x 3 + 7 x 4 = 0 ⇒ x 1 = s − 7 t , x 2 + 2 x 3 − 3 x 4 = 0 ⇒ x 2 = − 2 s + 3 t . So
x = s ( 1 − 2 1 0 ) + t ( − 7 3 0 1 ) . \mathbf x=s\begin{pmatrix}1\\-2\\1\\0\end{pmatrix}+t\begin{pmatrix}-7\\3\\0\\1\end{pmatrix}. x = s 1 − 2 1 0 + t − 7 3 0 1 . Kernel ( A ) = span { ( 1 , − 2 , 1 , 0 ) T , ( − 7 , 3 , 0 , 1 ) T } , dim = 2. \text{Kernel}(A)=\operatorname{span}\left\{(1,-2,1,0)^T,\ (-7,3,0,1)^T\right\},\qquad \dim=2. Kernel ( A ) = span { ( 1 , − 2 , 1 , 0 ) T , ( − 7 , 3 , 0 , 1 ) T } , dim = 2. Step 4 — Summary
Answer
dim Im A = 2 , basis { ( 1 , 1 , 3 ) T , ( 2 , 3 , 8 ) T } ; dim ker A = 2 , basis { ( 1 , − 2 , 1 , 0 ) T , ( − 7 , 3 , 0 , 1 ) T } . \boxed{\;\dim\operatorname{Im}A=2,\ \text{basis }\{(1,1,3)^T,(2,3,8)^T\};\quad \dim\ker A=2,\ \text{basis }\{(1,-2,1,0)^T,(-7,3,0,1)^T\}.\;} dim Im A = 2 , basis {( 1 , 1 , 3 ) T , ( 2 , 3 , 8 ) T } ; dim ker A = 2 , basis {( 1 , − 2 , 1 , 0 ) T , ( − 7 , 3 , 0 , 1 ) T } .