Let S=A=pmatrixa&b&dpmatrix: a,b,c,d0,1,2,3,4, A^2-4A+3I=O.

Question

Let $S=\{A=\begin{pmatrix}a&b\c&d\end{pmatrix}: a,b,c,d\in\{0,1,2,3,4\},\ A^2-4A+3I=O\}$. Then the number of matrices in $S$ for which the sum of the diagonal elements is equal to 4 is:

Accepted Answer

$A^2-4A+3I=(A-I)(A-3I)=0$, so $A$ has eigenvalues among $\{1,3\}$. Given $\operatorname{tr}(A)=a+d=4$, the eigenvalues must be exactly $1$ and $3$ (distinct), so $\det(A)=1\cdot3=3$.

So $a+d=4$ and $ad-bc=3$, i.e. $bc=a(4-a)-3$, with $a,b,c,d\in\{0,1,2,3,4\}$.

- $a=0,d=4$: $bc=-3$ --- not possible.
- $a=1,d=3$: $bc=0$. Ordered pairs $(b,c)$ with $b,c\in\{0,\dots,4\}$ and $bc=0$: $5+5-1=9$.
- $a=2,d=2$: $bc=1$. Only $(b,c)=(1,1)$: $1$ way.
- $a=3,d=1$: $bc=0$: $9$ ways.
- $a=4,d=0$: $bc=-3$ --- not possible.

Total $=9+1+9=19$.

Solutions · 1