ISI2016-MMA-18

Question

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

• A. $B^2 =1$
• B. $B^2 =0$
• C. $B^2 =A$
• D. $B^2 =B$

Answer 1

we know from sum of powers is

$$\sum_{k=0}^{n}z^k = 1 + z + z^2 +... +z^n = \frac{z^{(n+1)} -1}{z-1}$$

so for a matrix A, we can similarly define

$$\sum_{k=0}^{n}A^k = 1 + A + A^2 +... +A^n = (A - I)^{-1}(A^{n+1} - I)$$

such that $(A - I)^{-1}$ exists that is   $det(A - I) \neq 0$

as given in Question

$$B = A + A^2 +... +A^{50} = (A - I)^{-1}(A^{50+1} - I) - I $$

given,

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

we find,

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

now, calculating the determinant of (A-I)

$$ det(A-I) = det(\begin{bmatrix} -2 & -2\\ 0 & -2 \end{bmatrix}) = 4$$

Now we find the inverse of (A-I),

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

Computing $A^{51}$ by hand is tiresome so we need to check for patterns

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

and

$$ A^3 = A^2A \\ \\ A^3 = \begin{bmatrix} 1 & -4\\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 2\\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 6\\ 0 & -1 \end{bmatrix} $$

Now we see a Pattern, so therefore

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

Now we need to put everything together

$$ A^{51} = \begin{bmatrix} -1 & 102\\ 0 & -1 \end{bmatrix} $$

$$ B = (A - I)^{-1}(A^{50+1} - I) - I $$

$$ \implies B = \begin{bmatrix} -1/2 & -1/2\\ 0 & -1/2 \end{bmatrix} \left ( \begin{bmatrix} -1 & 102\\ 0 & -1 \end{bmatrix} - \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \right ) - \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} $$

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

$$ \implies B^2 = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} $$

Therefore Option B is correct Answer