Question

Let $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ be a matrix. If $A^6 = \begin{pmatrix} p & q \\ r & s \end{pmatrix}$, then

• A. number of factors common to 'p' and 's' is 1.
• B. number of factors of (p+q) is 7.
• C. (p+q+r+s) is an even integer.
• D. the sum of the proper divisors of (p+q+r+s) is 1820.

Answer 1

Answer: (A), (B), (C)

(A), (B), (C)

Answer 2

First, we calculate $A^6$. We observe a pattern in the powers of A:

$A^n = \begin{pmatrix} 2^n & 3^n - 2^n \\ 0 & 3^n \end{pmatrix}$.

Thus, $A^6 = \begin{pmatrix} 2^6 & 3^6 - 2^6 \\ 0 & 3^6 \end{pmatrix} = \begin{pmatrix} 64 & 665 \\ 0 & 729 \end{pmatrix}$.

So, $p=64$, $q=665$, $r=0$, $s=729$.

(A) The factors of $p=64=2^6$ are powers of 2, and the factors of $s=729=3^6$ are powers of 3. The only common factor is 1.

(B) $p+q = 64+665 = 729 = 3^6$. The number of factors of $3^6$ is $6+1 = 7$.

(C) $p+q+r+s = 64+665+0+729 = 1458$, which is an even integer.

(D) $p+q+r+s = 1458 = 2 \times 3^6$. The sum of all divisors of 1458 is $\sigma(1458) = (1+2)(1+3+3^2+3^3+3^4+3^5+3^6) = 3 \times 1093 = 3279$. The sum of the proper divisors is $3279 - 1458 = 1821$, not 1820.

Therefore, options (A), (B), and (C) are correct.