Do you know its property ?-13

Given : $A=\begin{pmatrix} 4 & 1 \\ -9 & -2 \end{pmatrix}=\begin{pmatrix} 1+3 & 1 \\ -9 & 1-3 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}=I + B$

I is the identity or unit matrix and lets me assume that $\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}=B$

Now $B^2=\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix} \times \begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

As I and B commute - unit matrix commutes with any matrix .

So using binomial expansion, we can write $A^{100}=\left( I +B \right)^{100}$

$\implies A^{100}=I+ ^{100}C_1.B + ^{100}C_2.B^2+....+^{100}C_{100}.B^{100}$

But $B^k=0, \forall k \geq 2$ . So the above reduces to

$A^{100}=I+100.B$

$\quad \quad =\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}+100.\begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$

$\quad \quad =\begin{pmatrix} 301 & 100 \\ -900 & -299 \end{pmatrix}$

$\implies Required=301+100-900-299=\boxed{-798}$

where $^nC_r=\dfrac{n!}{r! \times (n-r)!}$

enjoy

$A=\begin{pmatrix} 4 & 1 \\ -9 & -2 \end{pmatrix}$ .

$A^2=\begin{pmatrix} 7 & 2 \\ -18 & -5 \end{pmatrix}$ .

Now, using Induction we can prove that,

$A^n=\begin{pmatrix} 3n+1 & n \\ -9n & -3n+1 \end{pmatrix}$ .

Now, $A^{100}=\begin{pmatrix} 301 & 100 \\ -900 & -299 \end{pmatrix}$ .

So, $a+b+c+d=\boxed{-798}$

The general procedure for taking any matrix power is to find a similar matrix into the form ${P}^{-1}DP$ , where $D$ is the diagonal matrix. However, this matrix cannot be diagonalized due to the repeated eigenvalue $\lambda =1$ . Thus we must find the similar matrix in the Jordan form .

Here we get $A= \begin{pmatrix} -1 & -1/3 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} 1& 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0& 1/3 \\ -3 & -1 \end{pmatrix}$

hence

${A}^{100}= \begin{pmatrix} -1 & -1/3 \\ 3 & 0 \end{pmatrix} {\begin{pmatrix} 1& 1 \\ 0 & 1 \end{pmatrix}}^{100} \begin{pmatrix} 0& 1/3 \\ -3 & -1 \end{pmatrix}$

What is cool about the middle matrix is the property ${\begin{pmatrix} 1& 1 \\ 0 & 1 \end{pmatrix}}^{n} = \begin{pmatrix} 1& n \\ 0 & 1 \end{pmatrix}$

We multiply back the matrix and get ${A}^{100}= \begin{pmatrix} 301 & 100 \\ -900 & -299 \end{pmatrix}$