Question
Let $\text{S} = \left\{\text{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4\} \text{ and } \text{A}^2 - 4\text{A} + 3\text{I} = 0 \right\}$ 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:
The correct answer is
19
We are given a set of \(2 \times 2\) matrices \( \text{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) with elements \( a, b, c, d \in \{0, 1, 2, 3, 4\} \), satisfying the equation:
\(A^2 - 4A + 3I = 0\)
We need to find the number of matrices in this set for which the sum of the diagonal elements (\(a + d\)) is equal to 4.
Step-by-Step Solution:
- Substitute \(\text{A}\) and calculate \(\text{A}^2\):
- \(\text{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
- \(\text{A}^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{bmatrix}\)
- Given equation is:
- This simplifies to two equations:
- \(a^2 + bc - 4a + 3 = 0\)
- \(ab + bd - 4b = 0\)
- \(ac + cd - 4c = 0\)
- \(bc + d^2 - 4d + 3 = 0\)
- Considering the condition \(a + d = 4\).
- Substitute \(d = 4 - a\) into the equations as follows:
- \[a^2 + bc - 4a + 3 = 0 \quad \text{and} \quad bc + (4-a)^2 - 4(4-a) + 3 = 0\]
- This condition significantly reduces the possibilities. Try possible integer values:
- Values for \(a\) and \(d = 4-a\) are:
- \(a = 0, d = 4\)
- \(a = 1, d = 3\)
- \(a = 2, d = 2\)
- \(a = 3, d = 1\)
- \(a = 4, d = 0\)
- For each, solve for \(b\) and \(c\) using the equations above:
- After verifying each case, we find a total of 19 matrices satisfy these conditions.
Hence, the number of matrices in the set \(S\) for which the sum of the diagonal elements is 4 is 19.
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