a) Prove that (2 ^ 1 x) = 2x 1 x^2 . [2] b) Using the principle of mathematical induction, prove that: 2 + 2^2 + 2^3 + + 2^n = 2(2^n 1) . [2] c) If A = pmatrix 2 & 1 \\ 1 & 2 pmatrix , find AA^T . [2]

a) Prove that $\sin (2{\sin }^{-1}x)=2x\sqrt{1-x^2}$

Proof:

Let $\theta ={\sin }^{-1}x$. Then, by definition, $\sin \theta =x$.

We need to prove: $$\sin (2{\sin }^{-1}x)=2x\sqrt{1-x^2}$$

Substitute $\theta$ into the left-hand side (LHS): $$\sin (2\theta )=2\sin \theta \cos \theta$$

We know $\sin \theta =x$. To find $\cos \theta$, use the Pythagorean identity: $${\sin }^2\theta +{\cos }^2\theta =1⟹{\cos }^2\theta =1-{\sin }^2\theta =1-x^2$$ $$\cos \theta =\sqrt{1-x^2}(\text{since }\theta ={\sin }^{-1}x⟹\theta \in [-\pi /2,\pi /2],\text{ so }\cos \theta \ge 0)$$

Substitute $\sin \theta$ and $\cos \theta$ back into the expression for $\sin (2\theta )$: $$\sin (2\theta )=2\cdot x\cdot \sqrt{1-x^2}=2x\sqrt{1-x^2}$$

Thus, we have shown: $$\sin (2{\sin }^{-1}x)=2x\sqrt{1-x^2}$$

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b) Using the principle of mathematical induction, prove that:

$$2+2^2+2^3+\cdots +2^n=2(2^n-1)$$

Proof:

We will prove the statement by mathematical induction.

Base Case (n = 1):

For $n=1$, the left-hand side (LHS) is: $$2=2^1$$ The right-hand side (RHS) is: $$2(2^1-1)=2(2-1)=2$$ Thus, LHS = RHS, and the statement holds for $n=1$.

Inductive Step:

Assume the statement is true for some $k\ge 1$, i.e., $$2+2^2+2^3+\cdots +2^k=2(2^k-1)$$

We need to show that the statement holds for $k+1$, i.e., $$2+2^2+2^3+\cdots +2^k+2^{k+1}=2(2^{k+1}-1)$$

Starting with the LHS for $k+1$: $$2+2^2+2^3+\cdots +2^k+2^{k+1}=\left[2+2^2+2^3+\cdots +2^k\right]+2^{k+1}$$ By the inductive hypothesis: $$=2(2^k-1)+2^{k+1}$$ Simplify the expression: $$=2\cdot 2^k-2+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2(2^{k+1}-1)$$

Thus, the statement holds for $k+1$.

By the principle of mathematical induction, the statement is true for all positive integers $n$.

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c) If $A=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$, find $A{A}^T$.

Solution:

First, find the transpose of matrix $A$, denoted as $A^T$: $$A^T={\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)}^T=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$$

Wait, this is incorrect. The transpose of a matrix is obtained by swapping rows and columns. So, for $A=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$, the transpose is: $$A^T={\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)}^T=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$$

Wait, this is a symmetric matrix, so $A^T=A$. However, let's verify the transpose properly:

Given: $$A=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$$ The transpose $A^T$ is: $$A^T=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$$

Wait, this is incorrect. The transpose of a matrix is obtained by flipping the matrix over its main diagonal, switching the row and column indices of the matrix. So, for $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, $A^T=\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)$.

Thus, for $A=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$, the transpose is: $$A^T=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$$

Wait, this is the same as $A$. This is because $A$ is a symmetric matrix. So, $A^T=A$.

Now, compute $A{A}^T$: $$A{A}^T=A\cdot A=\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)\cdot \left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)$$

Perform the matrix multiplication: $$A{A}^T=\left(\begin{array}{cc}(2\cdot 2+1\cdot 1)& (2\cdot 1+1\cdot (-2))\\ (1\cdot 2+(-2)\cdot 1)& (1\cdot 1+(-2)\cdot (-2))\end{array}\right)$$

Calculate each element: $$A{A}^T=\left(\begin{array}{cc}(4+1)& (2-2)\\ (2-2)& (1+4)\end{array}\right)=\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)$$

Thus, the final answer is: $$A{A}^T=\boxed{\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)}$$