Series Matrix

Algebra Level 3

$\large {\begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix}}$

Denote the matrix above as $A$ . Find the value of series below?

$I + 2A + 3A^2 + 4A^3 + \ldots$

For more problems see this

\begin{bmatrix} 4 & 1 \\ -4 & -3 \end{bmatrix}

\begin{bmatrix} 4 & -3 \\ -8 & 5 \end{bmatrix}

\begin{bmatrix} 5 & 2 \\ -8 & -3 \end{bmatrix}

\begin{bmatrix} 5 & 3 \\ -4 & -8 \end{bmatrix}

3 solutions

Vishwas Gajera
Mar 19, 2015

A²=0, A³=0......... Now remains only first tow terms

I + 2A

Hi dude !

Try $A^{2}$ .This way it'll look good .

Hover your cursor over the $A^{2}$ to get the required Latex code .

A Former Brilliant Member - 6 years, 2 months ago

Parag Zode
Mar 20, 2015

$I=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

So ,value of $I= 1$ .

And, value of matrix $A=2 . (-2) - (-4) . 1 =0$ .

So ,the series will be like $1 +2(0) + 3(0)^{2} +4(0)^{3} +....$ which gives the answer as $1$

Now, we have to find a matrix whose value is equal to be $1$

Hence the matrix $\begin{bmatrix} 5 & 2 \\ -8 & -3 \end{bmatrix}$ is the answer.

I think something is wrong in your solution. We don't do it in this manner.

Himanshu Goel - 5 years, 5 months ago

I actually tried something after doing the problem: since we know the Taylor series for $\frac{1}{(1-A)^{2}}$ is the desired summation, I went ahead and calculated that, taking $1 = I_{2}$ . I ended up with

$\begin{pmatrix} 1 & 0 \\ -10/9 & 1/9 \end{pmatrix}$

Jake Lai - 6 years, 2 months ago

Tristan Goodman
Jul 16, 2020

I first attempted to diagonalize the matrix in order to calculate its powers, however in attempting to do so, found that it only has an eigenvalue of 0, making it non-diagonalizable, but also indicating it is nilpotent, meaning it's square has all elements of zero. Therefore, all powers of the matrix from 2 and up all vanish, leaving only the first two elements of the sum.

Series Matrix

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