GATE BT 2022 • Q39 MCQ • 2 Marks

Section 1: Engineering Mathematics

Linear Algebra

Subtopic: Fundamentals of Linear Algebra

Question Statement:

A 2×2 matrix P has an eigenvalue 𝜆ଵ= 2 with eigenvector 𝑥ଵ= 1 0    and another eigenvalue 𝜆ଶ= 5, with eigenvector 𝑥ଶ= 1 1    . The matrix P is

Options:

  • (A) 2 0 0 5      
  • (B) 2 3 0 5      
  • (C) 1 1 0 1      
  • (D) 1 1 1 0      

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Official Solution & Concept Analysis (GATE BT 2022 Q.39)** **1. Governing Concept**: - **Topic**: Linear Algebra & Matrix Diagonalization ($P = S \Lambda S^{-1}$) - Given eigenvalues and eigenvectors: - $\lambda_1 = 2$ with eigenvector $x_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ - $\lambda_2 = 5$ with eigenvector $x_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ - Form modal matrix $S$ with eigenvectors as columns: $S = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ - Inverse of $S$: $\det(S) = 1(1) - 1(0) = 1 \implies S^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ - Diagonal eigenvalue matrix $\Lambda$: $\Lambda = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$ - Reconstructing matrix $P = S \Lambda S^{-1}$: $S \Lambda = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 0 & 5 \end{pmatrix}$ $P = (S \Lambda) S^{-1} = \begin{pmatrix} 2 & 5 \\ 0 & 5 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2(1) + 5(0) & 2(-1) + 5(1) \\ 0(1) + 5(0) & 0(-1) + 5(1) \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 0 & 5 \end{pmatrix}$ **2. Option Analysis**: - **Option (A) is INCORRECT**: Diagonal matrix $\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$, which has eigenvectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, not $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. - **Option (B) [2 3 / 0 5] is CORRECT**: Exactly matches matrix reconstruction $P = \begin{pmatrix} 2 & 3 \\ 0 & 5 \end{pmatrix}$. - **Option (C) is INCORRECT**: Has eigenvalues 1 and 1. - **Option (D) is INCORRECT**: Inverted components. **3. Final Answer**: - **Correct Choice**: **B**

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