Matrix Multiplication

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 $$1-100+3675$$

$$=3576$$

Given,

$$A=\begin{pmatrix}1&2\\-1&3\end{pmatrix}$$

First find $$A^2=A\cdot A$$

$$A^2=\begin{pmatrix}1&2\\-1&3\end{pmatrix}\begin{pmatrix}1&2\\-1&3\end{pmatrix}$$

First row first column : $$1(1)+2(-1)=1-2=-1$$

First row second column : $$1(2)+2(3)=2+6=8$$

Second row first column : $$(-1)(1)+3(-1)=-1-3=-4$$

Second row second column : $$(-1)(2)+3(3)=-2+9=7$$

Therefore, $$A^2=\begin{pmatrix}-1&8\\-4&7\end{pmatrix}$$

Now find $$4A$$

$$4A=4\begin{pmatrix}1&2\\-1&3\end{pmatrix}=\begin{pmatrix}4&8\\-4&12\end{pmatrix}$$

Now calculate $$A^2-4A$$

$$=\begin{pmatrix}-1&8\\-4&7\end{pmatrix}-\begin{pmatrix}4&8\\-4&12\end{pmatrix}$$

Subtract corresponding entries :

$$=\begin{pmatrix}-1-4&8-8\\-4-(-4)&7-12\end{pmatrix}$$

$$=\begin{pmatrix}-5&0\\0&-5\end{pmatrix}$$

Trace of a matrix is sum of diagonal elements.

Therefore,

$$\text{Trace}(A^2-4A)=-5+(-5)=-10$$

Final Answer :

$$-10$$