Question Statement:
For the matrix A = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, if the relation a + b = c + d holds and a, b, c, d ≠ 0,
then which one of the following statements about A is FALSE?
Options:
-
(A) $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is an eigenvector
-
(B) $\lambda$ = a + b is an eigenvalue
-
(C) $\lambda$ = d - b is an eigenvalue
-
(D) $\lambda$ = d + b is an eigenvalue
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.37)**
**1. Intuitive Physical Setup & Given Parameters**:
Consider the $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $a, b, c, d \neq 0$ and satisfying:
a + b = c + d = S
We are asked to identify which of the four given statements is **FALSE**.
**2. Step-by-Step Linear Algebra Derivation**:
1. **Row Sum Property**:
Notice that the sum of the elements in each row is identical:
- Row 1: $a + b = S$
- Row 2: $c + d = S$
Multiply matrix $A$ by the column vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$:
A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a + b \\ c + d \end{bmatrix} = \begin{bmatrix} S \\ S \end{bmatrix} = S \begin{bmatrix} 1 \\ 1 \end{bmatrix} = (a+b) \begin{bmatrix} 1 \\ 1 \end{bmatrix}
- This proves that $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is an **eigenvector** corresponding to eigenvalue $\lambda_1 = a + b$.
- Thus, Statement A and Statement B are **TRUE**.
2. **Find the Second Eigenvalue $\lambda_2$**:
- The trace of matrix $A$ is the sum of its eigenvalues:
\operatorname{Tr}(A) = a + d = \lambda_1 + \lambda_2
- Since $\lambda_1 = a + b$:
a + d = (a + b) + \lambda_2 \implies \lambda_2 = (a + d) - (a + b) = d - b
- Thus, $\lambda = d - b$ is indeed an eigenvalue of $A$. Statement C is **TRUE**.
3. **Evaluating Statement D**:
- Statement D claims that $\lambda = d + b$ is an eigenvalue.
- But we have uniquely found the two eigenvalues of the $2 \times 2$ matrix: $\lambda_1 = a + b$ and $\lambda_2 = d - b$.
- For $\lambda_2$ to equal $d + b$, we would require $d - b = d + b \implies 2b = 0 \implies b = 0$, which directly contradicts the problem condition that $a, b, c, d \neq 0$.
- Therefore, Statement D is **FALSE**.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) [1 1]^T is an eigenvector**: ❌ True statement. Verified by direct matrix-vector multiplication.
- **(B) $\lambda$ = a + b is an eigenvalue**: ❌ True statement. Direct consequence of constant row sums.
- **(C) $\lambda$ = d - b is an eigenvalue**: ❌ True statement. Derived from trace identity $\lambda_2 = \operatorname{Tr}(A) - \lambda_1 = a + d - (a + b) = d - b$.
- **(D) $\lambda$ = d + b is an eigenvalue**: ✅ **CORRECT (FALSE STATEMENT)**. Contradicts the trace theorem since $b \neq 0$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **D**
- **Exam Takeaway**: When row sums of a square matrix are constant, the all-ones vector is an eigenvector with eigenvalue equal to that row sum. The remaining eigenvalue is readily obtained from $\operatorname{Tr}(A) = \sum \lambda_i$.