Find all matrices $A$ that satisfy
$$A^2 - 4 A + 4I = \begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix}.$$
I've tried solving this but I can only get a single equation constraining $b$ and $c$, i.e. $b/c=3/5$. I find it difficult to solve for the rest. Can anybody give me a hint? Thanks
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Let $X=A-2I$, then we are solving $X^2=B$. Since $X$ is a $2 \times 2$ matrix, therefore by Cayley-Hamilton it satisfies $$X^2-(\text{tr}X)X+(\det X)I=0.$$ This implies $$(\text{tr}X)X=B+(\det X)I.$$ But $X^2=B^ \implies \det X = \pm (\det B)=\pm 3.$ Thus $$(\text{tr}X)X=B\pm 3I.$$ Let $X=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, then $$(a+d)\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix} 4\pm3 & 3 \\ 5 & 6\pm 3 \end{bmatrix}$$ From this we can conclude that $$(a+d)^2=(4 \pm 3)+(6 \pm 3)=16,4.$$ Thus $$\color{red}{\text{tr}X=\pm 4, \pm 2}$$ Now $$(\text{tr}X)X=B\pm 3I \implies \color{blue}{X=\frac{1}{\text{tr}X}(B\pm 3I)}$$
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Anurag A posted an answer but deleted it after I brought up some concerns. However, after some additional thought I think his argument can be completed.
First, diagonalise $B$ (the matrix on the RHS): $B=PDP^{-1}$. Now rewrite the equation: $$ (A-2I)^2=PDP^{-1}\\ (P^{-1}(A-2I)P)^2=D\\ X^2=D $$ At this point, as $B$ has positive eigenvalues it is tempting to conclude that $X=\sqrt{D}$ where $$ \sqrt{D}=\begin{bmatrix}\pm\sqrt{D_{11}} & 0\\0 & \pm\sqrt{D_{22}}\end{bmatrix} $$ However, this is not obvious since $X^2=D$ for some matrix $X$ does not imply that $X$ is diagonal; for example: $$ \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}^2=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix} $$ But we can still make this work. If $$ X=\begin{bmatrix}a & b\\c & d\end{bmatrix} $$ then $$ X^2=\begin{bmatrix}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix} $$ If either $b\neq0$ or $c\neq0$, we can conclude that $a+d=0$, so $a^2+bc=d^2+bc$. Therefore, were $X$ not diagonal, the diagonal entries of $D$ would have to be the same, but it turns out they are not. So $X$ must be diagonal and indeed we can proceed from $$ X=\sqrt{D} $$ for each of the $4$ options for $\sqrt{D}$.
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Call the matrix on the left as $B$, your equation becomes $$(A-2I)^2=B~~~(1),$$ where $$B= \begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix} ~~ \mbox{and let} ~~\sqrt{B}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ This means $$A=2I + \sqrt{B}==~~~(2)$$. Now you have to find $\sqrt{B}$, Let $$B=\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2=\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix}=\begin{pmatrix} 4 & 3 \\ 5 & 6 \end{pmatrix}~~~~(3)$$ Now solve for $a,b,c,d$ the set of four equations $$a^2+bc=4~~~(i),~ b(a+d)=3~~~(ii), ~c(a+d)=5~~~(iii),~ d^2+bc=6~~~(iv)~~~(4)$$ Let $c=k$ then from (ii) and (iii) get $b=3k/5.$ From (i) and (ii) have $a^2-d^2=-2 \Rightarrow (a-d)(a+d)=-2.$ Next use (ii) to get $a-d=-6k/5$ and from (iii) we have $(a+d)=5/k$. Solve these two to get $a=(5/k-6k/5)/2$. Put this $a$ and $b,c$ in (i) to get $$ (5/k-6k/3)^2/4+3k^2/5=4 ~~~~~~(5)$$ whose roots are $$k=\pm 5/2,~~ \pm \frac{5}{2 \sqrt{6}}~~~~~~~~~~(6)$$ Four $2 \times 2$ matrices are possible for $\sqrt{B}$. $$\sqrt{B}=\frac{\pm 1}{2} \begin{pmatrix} 1 & 3 \\ 5 & 3 \end{pmatrix} \mbox{and two more for other value of}~ k~~~~ (7)$$ The other two matrices do satisfy $Trace(\sqrt{B})^2=Trace(B)$ but they do not satisfy $(\det |\sqrt{B}|)^2=\det |B|$, hence they are rejected. from (2) you get $$A_1=\frac{1}{2}\begin{pmatrix} 5 & 3\\ 5 & 7 \end{pmatrix},~~~A_2=\frac{1}{2}\begin{pmatrix} 3 & -3 \\ -5 & 1 \end{pmatrix}.$$ It will be fun to check that these two matrices for $A$ will satisfy the original equation. Note that my answer though consistent will be different from that of the Answer of other methods. Also note that this question does not have a unique matrix as answer.
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