Question
Let $S = \left\{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}:a,b,c,d \in \{ 0,1,2,3,4\} \right.\ $ and $\left. \ A^{2} - 4\text{ }A + 3I = 0 \right\}$ 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:
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Solution
Sol. [Key Concept: $A^{2} - 4A + 3I = 0 \Rightarrow \ a + d = 4$ and $ad - bc = 3$; count the integer entries.]
a, b, c, d $\in \{ 0,1,2,3,4\}$
$$a + d = 4\ ad - bc = 3 $$Case-I $\ a = 0\ \text{ }d = 4$
$$0 - bc = 3 $$No solution
Case-II $\ a = 1\ \text{ }d = 3$
$$3 - bc = 3 $$$bc = 0\ \Rightarrow 9$ ways
Case-III $a = 2\ \text{ }d = 2$
$$4 - bc = 3 $$$bc = 1\ \Rightarrow 1$ ways
Case-IV $\ a = 3\ \text{ }d = 1$
$bc = 0\ \Rightarrow 9$ way
Case-V $a = 4\ \text{ }d = 0 \Rightarrow$ No Solution
Total 19 matrix
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