Question:medium
Let \(S = \left\{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4\} \text{ and } A^2 - 4A + 3I = 0 \right\}\) be a set of \(2 \times 2\) matrices. Then the number of matrices in \(S\), for which the sum of the diagonal elements is equal to 4, is:
Let \(S = \left\{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4\} \text{ and } A^2 - 4A + 3I = 0 \right\}\) be a set of \(2 \times 2\) matrices. Then the number of matrices in \(S\), for which the sum of the diagonal elements is equal to 4, is:
Updated On: Jun 6, 2026
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- 19
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
By the Cayley-Hamilton theorem, any \(2 \times 2\) matrix \(A\) satisfies its own characteristic equation:
\[ A^2 - (\text{Trace}(A))A + (\text{Det}(A))I = 0 \] We are given that \(A^2 - 4A + 3I = 0\) and the sum of the diagonal elements (Trace) is 4.
Step 2: Key Formula or Approach:
Since \(\text{Trace}(A) = a+d = 4\), the characteristic polynomial of \(A\) is exactly \(x^2 - 4x + (ad-bc) = 0\).
Comparing this with \(A^2 - 4A + 3I = 0\), we deduce that the determinant must be 3.
Thus, we have two conditions:
1. \(a + d = 4\)
2. \(ad - bc = 3\)
Step 3: Detailed Explanation:
From the first condition, \(d = 4 - a\).
Substitute \(d\) into the determinant condition:
\[ a(4 - a) - bc = 3 \implies 4a - a^2 - 3 = bc \implies -(a^2 - 4a + 3) = bc \] \[ \implies -(a - 1)(a - 3) = bc \] Since \(a, b, c, d \in \{0, 1, 2, 3, 4\}\), we must have \(bc \ge 0\).
Therefore, \(-(a - 1)(a - 3) \ge 0 \implies (a - 1)(a - 3) \le 0\).
This implies \(1 \le a \le 3\). Since \(a\) is an integer, \(a\) can be 1, 2, or 3.
Let's test each case:
Case 1: \(a = 1\)
Then \(d = 3\) and \(bc = -(1-1)(1-3) = 0\).
Since \(bc = 0\), either \(b = 0\) or \(c = 0\).
The pairs \((b, c)\) are:
- \(b = 0\), \(c \in \{0, 1, 2, 3, 4\}\) (5 pairs)
- \(c = 0\), \(b \in \{1, 2, 3, 4\}\) (4 pairs)
Total pairs for \(a=1\) is \(5 + 4 = 9\).
Case 2: \(a = 2\)
Then \(d = 2\) and \(bc = -(2-1)(2-3) = 1\).
Since \(b, c\) are non-negative integers, \(bc = 1 \implies b = 1\) and \(c = 1\).
Total pairs for \(a=2\) is \(1\).
Case 3: \(a = 3\)
Then \(d = 1\) and \(bc = -(3-1)(3-3) = 0\).
Similar to Case 1, this gives 9 pairs.
Step 4: Final Answer:
The total number of such matrices is \(9 + 1 + 9 = 19\).
By the Cayley-Hamilton theorem, any \(2 \times 2\) matrix \(A\) satisfies its own characteristic equation:
\[ A^2 - (\text{Trace}(A))A + (\text{Det}(A))I = 0 \] We are given that \(A^2 - 4A + 3I = 0\) and the sum of the diagonal elements (Trace) is 4.
Step 2: Key Formula or Approach:
Since \(\text{Trace}(A) = a+d = 4\), the characteristic polynomial of \(A\) is exactly \(x^2 - 4x + (ad-bc) = 0\).
Comparing this with \(A^2 - 4A + 3I = 0\), we deduce that the determinant must be 3.
Thus, we have two conditions:
1. \(a + d = 4\)
2. \(ad - bc = 3\)
Step 3: Detailed Explanation:
From the first condition, \(d = 4 - a\).
Substitute \(d\) into the determinant condition:
\[ a(4 - a) - bc = 3 \implies 4a - a^2 - 3 = bc \implies -(a^2 - 4a + 3) = bc \] \[ \implies -(a - 1)(a - 3) = bc \] Since \(a, b, c, d \in \{0, 1, 2, 3, 4\}\), we must have \(bc \ge 0\).
Therefore, \(-(a - 1)(a - 3) \ge 0 \implies (a - 1)(a - 3) \le 0\).
This implies \(1 \le a \le 3\). Since \(a\) is an integer, \(a\) can be 1, 2, or 3.
Let's test each case:
Case 1: \(a = 1\)
Then \(d = 3\) and \(bc = -(1-1)(1-3) = 0\).
Since \(bc = 0\), either \(b = 0\) or \(c = 0\).
The pairs \((b, c)\) are:
- \(b = 0\), \(c \in \{0, 1, 2, 3, 4\}\) (5 pairs)
- \(c = 0\), \(b \in \{1, 2, 3, 4\}\) (4 pairs)
Total pairs for \(a=1\) is \(5 + 4 = 9\).
Case 2: \(a = 2\)
Then \(d = 2\) and \(bc = -(2-1)(2-3) = 1\).
Since \(b, c\) are non-negative integers, \(bc = 1 \implies b = 1\) and \(c = 1\).
Total pairs for \(a=2\) is \(1\).
Case 3: \(a = 3\)
Then \(d = 1\) and \(bc = -(3-1)(3-3) = 0\).
Similar to Case 1, this gives 9 pairs.
Step 4: Final Answer:
The total number of such matrices is \(9 + 1 + 9 = 19\).
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