Powers of a Matrix

Solution :

Given,

$$A=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$$

First calculate

$$A^2$$

$$A^2=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$$

$$=\begin{bmatrix}7 & 10 \\15 & 22\end{bmatrix}$$

Now,
$$5A=\begin{bmatrix}5 & 10 \\15 & 20\end{bmatrix}$$

Therefore,

$$A^2-5A=\begin{bmatrix}7 & 10 \\15 & 22\end{bmatrix} - \begin{bmatrix}5 & 10 \\15 & 20\end{bmatrix}$$

$$=\begin{bmatrix}2 & 0 \\0 & 2\end{bmatrix}$$

Given,

$$A^2-5A+kI=0$$

Therefore,

$$2I+kI=0$$

$$(2+k)I=0$$

$$2+k=0$$

$$k=-2$$

Final Answer :

$$-2$$

Observe the pattern of powers of $$A.$$

We have $$A^2= \begin{bmatrix} 1&2&3\\ 0&1&2\\ 0&0&1 \end{bmatrix}$$ and $$A^3= \begin{bmatrix} 1&3&6\\ 0&1&3\\ 0&0&1 \end{bmatrix}$$

Hence, by induction,

$$A^n=\begin{bmatrix}1&n&\frac{n(n+1)}2\\0&1&n\\0&0&1\end{bmatrix}$$

Therefore,

$$A^{50}=\begin{bmatrix}1&50&1275\\0&1&50\\0&0&1\end{bmatrix}$$

Now find its inverse.

Using the standard inverse form for such upper triangular matrices,

$$A^{-n}=\begin{bmatrix}1&-n&\frac{n(n-1)}2\\0&1&-n\\0&0&1\end{bmatrix}$$

Thus,

$$A^{-50}= \begin{bmatrix} 1&-50&1225\\ 0&1&-50\\ 0&0&1 \end{bmatrix}$$

Given,

$$A^{50}B= \begin{bmatrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{bmatrix}$$

Hence,

$$B= A^{-50} \begin{bmatrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{bmatrix}$$

$$= \begin{bmatrix} 1&-50&1225\\ 0&1&-50\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{bmatrix}$$

$$= \begin{bmatrix} 1&-100&3675\\ 0&2&-150\\ 0&0&3 \end{bmatrix}$$

The first row of $$B$$ is $$[1,\,-100,\,3675]$$ Hence the required sum is $$1-100+3675$$

$$=3576$$