# ISI2016

857 views

Let $A$ be a matrix such that:

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

and $B=A+A^2+A^3+\ldots +A^{50}$. Then which of the following is true?

1. $B^{2}=I$
2. $B^{2}=0$
3. $B^{2}=B$
4. None of the above

edited

$B=\begin{pmatrix} {-1}&2 \\ 0&{-1}\end{pmatrix}+\begin{pmatrix}1&{-4} \\ 0&1\end{pmatrix}+\begin{pmatrix} {-1}&6\\ 0&{-1}\end{pmatrix}+\cdots$

$=\begin{pmatrix} 0&(2-4)+(6-8)\cdots \\ 0&0\end{pmatrix}$

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

$=\begin{pmatrix} 0&{-2}\times 25 \\ 0&0\end{pmatrix}$

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

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

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

edited
0
Shouldnt this be A ?

Can you elaborate :)
1
chk @Dexter :)
0
@srestha is A2 = $A^2$, A3 = $A^3$ ? how did you know this?
1
I just assumed, u can ans if any other way possible :)
1
no other way is possible :) I was thinking the same else how to know what are A2, A3 and so on..?
2
haha , that is why I assumed :)
0
you can also say think this way

when adding to "ODD' number of A'S result =1

when adding with an even number of A'S result = 0

Here it's upto A^50 (EVEN) then result will be ZERO
0
@srestha , given matrix A is not diagonalizable. right ?
0

https://gateoverflow.in/204100/gate2018-26

Can repeated eigen value means not diagonalizable? :(

1

@srestha

'repeated' eigenvalues means may or may not be diagonalizable matrix but 'distinct' eigen values  must guarantee that given matrix is diagonalizable.

because repeated eigen values may produce less than n independent eigen vectors  for a given $n \times n$ matrix. So , these type of matrices can't be diagonalised.

but it is not always the case. For example , Identity matrix 'I' of size $n \times n$  has repeated eigen values but still it is diagonal matrix and produces 'n' linearly independent vectors.

I have seen this gate18 question  but I have got doubt in this question that given matrix is diagonalizable or not because this question can also be solved by diagonalization concept to get the power of matrix. but I think it is giving only 1 linearly independent eigen vactor, so it can't be diagonalized. that's why I have asked whether given matrix is diagonalizable or not. :)

1
ex : identity matrix which has only 1 as eigen value but its trivial diagonalizable

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

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

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

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

$A^5=\begin{bmatrix} -1 & 10\\ 0 & -1 \end{bmatrix}$

so we can generalize $A^k=$$\begin{bmatrix} (-1)^k & 2k.(-1)^{k+1}\\ 0 & (-1)^k \end{bmatrix}$

Given matrix A=$\begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$

When we sum up $A+A^2+A^3+....A^{50}=B$ it is very clear that $b_{21}=0$ and since half of the matrix powers are odd and half are even, so when we add all of them together then $b_{11}=b_{22}=0$

Now, we need to compute the entry $b_{12}$

To compute it I take sum of all entries of $a_{12}$ in $A+A^3+A^5+...A^{49}$

So this will be like $2+6+10+....+98 = \frac{98+2}{2} *25 =1250$ and this will be positive in sign.

Now, I take the sum of all entries of $a_{12}$ in $A^2+A^4+....A^{50}$

and this is like $-4-8.....-100 = \frac{-104}{2} * 25=-1300$

So, $b_{12}$ entry would be $1250-1300=-50$

Hence our matrix B would be $\begin{bmatrix} 0 &-50\\ 0 & 0 \end{bmatrix}$

$B^2=0.$ option(B)

Note : To take sum of AP $a_1+a_2+a_3+...a_n= \frac{a_n(last\, term)+a_1 (first \, term)}{2}* \text{number of terms}$

edited

Convert stipulated matrix to Identity (like) matrix.

$R_{1} = R_{1} + 2 R_{2}$

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

Now every  even power of A will be positive I and every odd power will be negative I.

So they will cancel each other out and answer will be 0.

If I missed something, let me know.

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