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If $A = \begin{bmatrix}2&1\\-4&-2\end{bmatrix}$, then $I+2A+3A^2+4A^3+\dots$ is equal to

(a) $\begin{bmatrix}4&1\\-4&0\end{bmatrix}$

(b) $\begin{bmatrix}3&1\\-4&-1\end{bmatrix}$

(c) $\begin{bmatrix}5&2\\-8&-3\end{bmatrix}$

(d) $\begin{bmatrix}5&2\\-3&-8\end{bmatrix}$

How does one solve such equations formed by matrices?

Community
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mathuser121
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1 Answers1

1

$A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix} \implies A^n = \begin{bmatrix}0&0\\0&0\end{bmatrix} \space\forall\space n \ge2$

$I + 2A +3A^2 + 4A^3 +\dots=I+2A = \begin{bmatrix}5&2\\-8&-3\end{bmatrix}$

Maadhav
  • 1,557
  • @mathuser121 Hah, wouldn't say a genius but thanks! Firstly, I would say look up some problems on Brilliant.org, they have the best set of problems. Some books I would recommend would be Linear Algebra And its Applications. By Gilbert Strang, Linear Algebra and Its Applications David C. Lay. These are beginner-intermediate level. – Maadhav Dec 26 '17 at 09:13
  • Some more good references https://math.stackexchange.com/questions/160056/what-is-a-good-book-to-study-linear-algebra?noredirect=1&lq=1 https://math.stackexchange.com/questions/804716/very-good-linear-algebra-book?noredirect=1&lq=1 – Maadhav Dec 26 '17 at 09:16