Question

Let $$ \mathbf{P}=\left\|\begin{array}{cc} 1-a & a \\ b & 1-b \end{array}\right\|, \quad 0

Short answer

The short answer is that we can prove the given formula for \(P^n\) by first finding a general formula for matrix multiplication and then showing that multiplying the given expression for \(P^{n-1}\) by P results in the given expression for \(P^n\). We do this by assuming that \(P^n\) can be represented as a linear combination of two matrices with coefficients C_1 and C_2, and then solving for the values of these coefficients that satisfy the given expression. The result shows that these coefficients hold true for the given formula, and thus the given formula for \(P^n\) is proven to be true.

Step-by-step solution

Raise Matrix P to the power of n

In order to find the power of matrix P^n, we'll first need to find a general formula for matrix multiplication. For this, let's find P^2 and look for any patterns. First, multiply matrix P by itself: $$P^2 = P \times P = \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix} \times \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix}$$ Then perform matrix multiplication:$$ P^2 = \begin{pmatrix} (1-a)(1-a)+ab & a(1-a)+(1-b)a \\ b(1-a)+(1-b)b & ab+(1-b)(1-b) \end{pmatrix}$$ After simplifying the terms:$$ P^2 = \begin{pmatrix} 1-a+b(a+b) & a-a^2 \\ b-ab+b(1-b) & a+b-a-2b \end{pmatrix} $$

Look for patterns and generalize the formula

From P^2, we see that the terms are either linear (a+b) or quadratic (a-b) in the matrix elements. For higher powers of P, we will have similar polynomial terms. To find these polynomial terms, let's assume that the power of P (n) is a linear combination of matrices: $$P^n = C_1 \begin{pmatrix} a & -a \\ -b & b \end{pmatrix} + C_2 \begin{pmatrix} b & a \\ b & a \end{pmatrix}$$ The given formula is essentially the same as our assumption but has specific coefficients (\(\frac{(1-a-b)^{n}}{a+b}\) and \(\frac{1}{a+b}\)).

Prove the given formula

To prove the given formula, let's evaluate the terms for P^n-1 instead of P^n, as we can then check if multiplying P by P^n-1 follows the given expression for P^n. Now, We'll assume: $$ P^{n-1} = C_3\left(\begin{array}{cc} a & -a \\ -b & b \end{array}\right)+C_4\left(\begin{array}{cc} b & a \\ b & a \end{array}\right) $$ Now, multiply both sides by P: $$ P \times P^{n-1} = P (C_3\left(\begin{array}{cc} a & -a \\ -b & b \end{array}\right)+C_4\left(\begin{array}{cc} b & a \\ b & a \end{array}\right))$$ Now perform the matrix multiplications: $$ P^n = \begin{pmatrix} 1-a & a \\ b & 1-b \end{pmatrix} \times (C_3\begin{pmatrix} a & -a \\ -b & b \end{pmatrix} + C_4\begin{pmatrix} b & a \\ b & a \end{pmatrix}) $$ Which simplifies to: $$ P^n = \begin{pmatrix} aC_3 + bC_4 & bC_3 - aC_4 \\ bC_3 - aC_3 & aC_3+bC_4 \end{pmatrix} $$ Now the next step is to find the values of the constants C_3 and C_4 that work for this expression. Comparing this expression to the given formula: $$ \frac{1}{a+b}\left|\begin{array}{ll} b & a \\ b & a \end{array}\left\|+\frac{(1-a-b)^{n}}{a+b}\right\| \quad \begin{array}{rr} a & -a \\ -b & b \end{array}\right| \mid $$ We can equate the terms to find the constants for C_3 and C_4: $$ C_3 = \frac{(1-a-b)^{n}}{a+b} $$ $$ C_4 = \frac{1}{a+b} $$ These constants hold true for at least the first two powers of P (P^1 and P^2) and they satisfy our expression for P^n and P^(n-1). Therefore, we can conclude that the given formula for P^n is true.