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UPSC 2019 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Matrix of a linear transformation · Linear Algebra · asked 10× in 13 yrs · Read the full method →

Question

Let T:R2R2T:\mathbb R^2\to\mathbb R^2 be a linear map such that T(2,1)=(5,7)T(2,1)=(5,7) and T(1,2)=(3,3)T(1,2)=(3,3). If AA is the matrix corresponding to TT with respect to the standard bases e1,e2e_1,e_2, then find Rank (A)(A).

Technique

Recover the standard matrix via A=[images]P1A=[\text{images}]\,P^{-1}; rank from a nonzero determinant. Faster: image of a basis is independent \Rightarrow full rank.

Solution

Step 1 — Set up the equation for AA

Let A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix} be the standard matrix of TT, so T(v)=AvT(v)=Av (columns). The data say

A[21]=[57],A[12]=[33].A\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}5\\7\end{bmatrix},\qquad A\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}3\\3\end{bmatrix}.

Stacking these as columns,

A[2112]=[5373].A\begin{bmatrix}2&1\\1&2\end{bmatrix}=\begin{bmatrix}5&3\\7&3\end{bmatrix}.

Step 2 — Solve for AA

Let P=[2112]P=\begin{bmatrix}2&1\\1&2\end{bmatrix}, detP=41=30\det P=4-1=3\neq0, so PP is invertible and

P1=13[2112].P^{-1}=\frac{1}{3}\begin{bmatrix}2&-1\\-1&2\end{bmatrix}.

Then

A=[5373]P1=13[5373][2112]=13[1035+61437+6]=13[71111].A=\begin{bmatrix}5&3\\7&3\end{bmatrix}P^{-1}=\frac{1}{3}\begin{bmatrix}5&3\\7&3\end{bmatrix}\begin{bmatrix}2&-1\\-1&2\end{bmatrix}=\frac{1}{3}\begin{bmatrix}10-3&-5+6\\14-3&-7+6\end{bmatrix}=\frac{1}{3}\begin{bmatrix}7&1\\11&-1\end{bmatrix}. A=[7/31/311/31/3].A=\begin{bmatrix}7/3&1/3\\[1mm]11/3&-1/3\end{bmatrix}.

Step 3 — Rank of AA

detA=19(7(1)111)=7119=189=20.\det A=\frac{1}{9}\big(7\cdot(-1)-1\cdot 11\big)=\frac{-7-11}{9}=\frac{-18}{9}=-2\neq0.

A 2×22\times2 matrix with nonzero determinant is invertible, hence of full rank.

Answer

  Rank(A)=2.  \boxed{\;\operatorname{Rank}(A)=2.\;}
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