Question

If $S = \{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, A^2 - 4A + 3I = \text{Null matrix}, a, b, c, d \in \{0, 1, 2, 3, 4\} \text{ and } \text{Tr}(A) = 4 \}$. Then the number of elements of set $S$ is/are

• A. 19
• B. 18
• C. 17
• D. 15

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By Cayley-Hamilton, for a $2 \times 2$ matrix: $A^2 - \text{Tr}(A)A + |A|I = 0$.
Comparing with $A^2 - 4A + 3I = 0$:
$\text{Tr}(A) = a+d = 4$ and $|A| = ad - bc = 3$.
Step 2: Detailed Explanation:
1. $a, d \in \{0, 1, 2, 3, 4\}$ and $a+d = 4$.
Pairs $(a, d)$: $(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)$.
2. Condition $bc = ad - 3$:
- $(0, 4) \implies bc = -3$ (Not possible).
- $(4, 0) \implies bc = -3$ (Not possible).
- $(1, 3) \implies bc = 3-3 = 0 \implies b=0$ or $c=0$.
$b=0 \implies c \in \{0, 1, 2, 3, 4\}$ (5 cases).
$c=0 \implies b \in \{0, 1, 2, 3, 4\}$ (5 cases).
Total $5+5-1 = 9$ matrices.
- $(3, 1) \implies bc = 0 \implies 9$ matrices (same as above).
- $(2, 2) \implies bc = 4-3 = 1 \implies b=1, c=1$ (1 matrix).
3. Total matrices $= 9 + 9 + 1 = 19$.
Step 4: Final Answer:
There are 19 elements in set $S$.