Question

Let A be a square matrix satisfying $A^2=2I$ and $\sum_{k=1}^{2n} A^k=63A(A+I)$, then the value of n

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (B)

6

Answer 2

Given the matrix equation $A^2 = 2I$. We need to evaluate the sum $\sum_{k=1}^{2n} A^k$. The powers of $A$ can be determined as follows: $A^1 = A$ $A^2 = 2I$ $A^3 = A^2 \cdot A = 2I \cdot A = 2A$ $A^4 = A^2 \cdot A^2 = (2I)(2I) = 4I$ $A^5 = A^4 \cdot A = 4I \cdot A = 4A$ $A^6 = A^4 \cdot A^2 = (4I)(2I) = 8I$

In general, for odd $k$, $A^k = 2^{(k-1)/2}A$. For even $k$, $A^k = 2^{k/2}I$.

Now, let's split the sum into odd and even powers: $\sum_{k=1}^{2n} A^k = (A^1 + A^3 + \dots + A^{2n-1}) + (A^2 + A^4 + \dots + A^{2n})$

Sum of odd powers: $A^1 + A^3 + \dots + A^{2n-1} = A + 2A + 4A + \dots + 2^{n-1}A$ $= A(1 + 2 + 4 + \dots + 2^{n-1})$ This is a geometric series with first term $1$, ratio $2$, and $n$ terms. The sum is $1 \cdot \frac{2^n - 1}{2 - 1} = 2^n - 1$. So, the sum of odd powers is $A(2^n - 1)$.

Sum of even powers: $A^2 + A^4 + \dots + A^{2n} = 2I + 4I + 8I + \dots + 2^nI$ $= I(2 + 4 + 8 + \dots + 2^n)$ This is a geometric series with first term $2$, ratio $2$, and $n$ terms. The sum is $2 \cdot \frac{2^n - 1}{2 - 1} = 2(2^n - 1)$. So, the sum of even powers is $2I(2^n - 1)$.

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

Now consider the right side of the given equation: $63A(A+I)$. $63A(A+I) = 63(A^2 + A)$ Since $A^2 = 2I$, we substitute this into the expression: $63(2I + A) = 63(A + 2I)$.

Equating the left and right sides of the given equation: $(2^n - 1)(A + 2I) = 63(A + 2I)$.

To solve for $n$, we need to consider if $(A+2I)$ is invertible. The eigenvalues of $A$ satisfy $\lambda^2 = 2$, so $\lambda = \pm\sqrt{2}$. The eigenvalues of $A+2I$ are $\lambda + 2$, which are $\sqrt{2}+2$ and $-\sqrt{2}+2$. Since these eigenvalues are non-zero, the matrix $A+2I$ is invertible.

We can divide both sides of the equation by $(A+2I)$: $2^n - 1 = 63$ $2^n = 63 + 1$ $2^n = 64$ $2^n = 2^6$ Therefore, $n = 6$.

The value of $n$ is 6.