Matrices

Question

Let $\begin{pmatrix} -1 &2 \\ 0 & -1 \end{pmatrix}$,and $B=A+A^{2}+A^{3}+\dots +A^{50}$, then

• A. $B^{2}=I$            
• B. $B^{2}=0$              
• C. $B^{2}=A$               
• D. $B^{2}=B$

Answer 1

$A=\begin{pmatrix} -1 &2 \\ 0& -1 \end{pmatrix}$

$A^{2}=\begin{pmatrix} -1 &2 \\ 0& -1 \end{pmatrix}\begin{pmatrix} -1 &2 \\ 0& -1 \end{pmatrix}=\begin{pmatrix} 1 &-4 \\ 0& 1 \end{pmatrix}$

$A^{3}=\begin{pmatrix} -1 &6 \\ 0& -1 \end{pmatrix}$

$A^{4}=\begin{pmatrix} 1 &-8 \\ 0& 1 \end{pmatrix}$

Given,$B=A+A^{2}+A^{3}+......+A^{50}$

$B=\begin{pmatrix} -1+1-1+1....+1 \ \ \ \ \ \ \ &2-4+6-8+10+.......-100 \\ 0& -1+1-1+1....+1 \end{pmatrix}$

$B=\begin{pmatrix} 0 \ \ &2-4+6-8+10+.......-100 \\ 0& 0 \end{pmatrix}$

Let $X=2-4+6-8+10+.......-100$

We can divide it into 2 series

$2+6+10+...98$      and $-4-8-12.......-94-100$

Let's take $2+6+10+...98$  

This sequence is an Arithmetic Progression.

Sum of Arithmetic Progression=$\frac{n}{2}[2a+(n-1)d]$

Here a=2 , d=4 , n=25

$\frac{25}{2}[4+24 \times 4]$

$=1250$

Now take $-4-8-12.......-94-100$

$-4-8-12.......-94-100$ = $-4(1+2+3+....+25)$

                                 =$-4 \times\frac{25 \times 26}{2}$

                                  =$-1300$

 $X=2-4+6-8+10+.......-100$

$X=2+6+10+...+98-4-8-12...-100 \\ X=1250-1300 \\ X=-50$

Put the value of X into the matrix

$B=\begin{pmatrix} 0 &-50 \\ 0 & 0 \end{pmatrix}$

$B^{2}=\begin{pmatrix} 0 &-50 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 &-50 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 &0 \\ 0 & 0 \end{pmatrix}$

$B^{2}=0$

Hence,Option$(B)$$B^{2}=0$ is the correct choice.

Comments

(2-4)+(6-8)....(98-100)=-2-2...-2 = -50 because such pairs are 25.