Question

If A is a square matrix and $e^A$ is defined as $e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + ....$ upto infinity = $\frac{1}{2} \begin{bmatrix} f(x) & g(x) \\ g(x) & f(x) \end{bmatrix}$, where A = $\begin{bmatrix} X & X \\ X & X \end{bmatrix}$, and I being the identity matrix of same order as A then :

If $\int \frac{g(x)}{f(x)} dx = log(e^{\alpha x} + 1) + \beta x + C$ then value of $(12\alpha + \beta)$ is : (where C is constant of integration)

Answer 1

23

Answer 2

Let the given matrix be $A = \begin{bmatrix} X & X \\ X & X \end{bmatrix}$. We need to compute $e^A$.

The definition of $e^A$ is given by the matrix exponential series: $e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + ....$

Let's calculate the powers of $A$. $A^2 = A \cdot A = \begin{bmatrix} X & X \\ X & X \end{bmatrix} \begin{bmatrix} X & X \\ X & X \end{bmatrix} = \begin{bmatrix} X^2 + X^2 & X^2 + X^2 \\ X^2 + X^2 & X^2 + X^2 \end{bmatrix} = \begin{bmatrix} 2X^2 & 2X^2 \\ 2X^2 & 2X^2 \end{bmatrix} = 2X^2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$. Let $A_0 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$. Then $A = X A_0$. $A^2 = 2X^2 A_0$. $A^3 = A^2 A = (2X^2 A_0)(X A_0) = 2X^3 A_0^2$. Let's compute $A_0^2$: $A_0^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = 2A_0$. So, $A^3 = 2X^3 (2A_0) = 4X^3 A_0 = 2^2 X^3 A_0$. $A^4 = A^3 A = (2^2 X^3 A_0)(X A_0) = 2^2 X^4 A_0^2 = 2^2 X^4 (2A_0) = 2^3 X^4 A_0$. In general, for $n \ge 1$, $A^n = 2^{n-1} X^n A_0 = 2^{n-1} X^n \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.

Now substitute this into the series for $e^A$: $e^A = I + \sum_{n=1}^{\infty} \frac{A^n}{n!} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \sum_{n=1}^{\infty} \frac{2^{n-1} X^n}{n!} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ $e^A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{2^n X^n}{n!} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{(2X)^n}{n!} \right) \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$. The series $\sum_{n=1}^{\infty} \frac{z^n}{n!} = e^z - 1$. Let $z = 2X$. So, $\sum_{n=1}^{\infty} \frac{(2X)^n}{n!} = e^{2X} - 1$. $e^A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \frac{1}{2} (e^{2X} - 1) \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} \frac{e^{2X} - 1}{2} & \frac{e^{2X} - 1}{2} \\ \frac{e^{2X} - 1}{2} & \frac{e^{2X} - 1}{2} \end{bmatrix}$ $e^A = \begin{bmatrix} 1 + \frac{e^{2X} - 1}{2} & \frac{e^{2X} - 1}{2} \\ \frac{e^{2X} - 1}{2} & 1 + \frac{e^{2X} - 1}{2} \end{bmatrix} = \begin{bmatrix} \frac{2 + e^{2X} - 1}{2} & \frac{e^{2X} - 1}{2} \\ \frac{e^{2X} - 1}{2} & \frac{2 + e^{2X} - 1}{2} \end{bmatrix} = \begin{bmatrix} \frac{e^{2X} + 1}{2} & \frac{e^{2X} - 1}{2} \\ \frac{e^{2X} - 1}{2} & \frac{e^{2X} + 1}{2} \end{bmatrix}$.

We are given that $e^A = \frac{1}{2} \begin{bmatrix} f(x) & g(x) \\ g(x) & f(x) \end{bmatrix}$. Comparing the elements, assuming the variable is $x$ instead of $X$: $\frac{1}{2} f(x) = \frac{e^{2x} + 1}{2} \implies f(x) = e^{2x} + 1$. $\frac{1}{2} g(x) = \frac{e^{2x} - 1}{2} \implies g(x) = e^{2x} - 1$.

We need to evaluate the integral $\int \frac{g(x)}{f(x)} dx$. $\int \frac{e^{2x} - 1}{e^{2x} + 1} dx$. We can rewrite the integrand by dividing the numerator and denominator by $e^x$: $\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. Let $u = e^x + e^{-x}$. Then $du = (e^x - e^{-x}) dx$. The integral becomes $\int \frac{du}{u} = \log|u| + C = \log|e^x + e^{-x}| + C$. Since $e^x + e^{-x} > 0$ for all real $x$, we have $\log(e^x + e^{-x}) + C$.

We need to express this in the form $log(e^{\alpha x} + 1) + \beta x + C$. $\log(e^x + e^{-x}) = \log(e^{-x}(e^{2x} + 1)) = \log(e^{-x}) + \log(e^{2x} + 1) = -x + \log(e^{2x} + 1)$. So, $\int \frac{g(x)}{f(x)} dx = \log(e^{2x} + 1) - x + C$.

Comparing this with the given form $log(e^{\alpha x} + 1) + \beta x + C$, we identify the coefficients: $\alpha = 2$ $\beta = -1$

We are asked to find the value of $(12\alpha + \beta)$. $12\alpha + \beta = 12(2) + (-1) = 24 - 1 = 23$.

The final answer is $\boxed{23}$.