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- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{alig... \implies \cos (180-60) & = (-1)\left(\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- lign*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}}\\ &= \sqrt{\frac{... 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}.... & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25} \right)\\ \end{ali
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following for... \[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\] \[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 y_2 - x_2 y_1}{x_1^2 + y
- Question 1, Exercise 2.5 @math-11-nbf:sol:unit02
- }1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left... }1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25 \end{array}\right]\quad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{... 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25 \end{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \text{R} \left[\
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- i}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right) = -\dfrac{\pi}{4} \\ \implies & \dfrac{y}{x-1} = \tan\left(-\dfrac{\pi}{4}\right) \\ \implies & \dfrac{y}{x-1} = -1 \\ \implies & ... in Cartesian form: $z \bar{z}=4\left|e^{i \theta}\right|$ ** Solution. ** Suppose $z=x+iy$, then $\bar{... As \begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| \\ \implies & (x+iy)(x-iy)=4|\cos\theta+i\sin\th
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{array}\right]$ if it exists. Also verify your answer by showin... -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ |A|&=0+1(-4)-1(1-3)\\ &=-4+3\\ &=-1\neq 0\end{... & 1 & 0 \\ 1 & -1 & 4 & 0 & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{array}{ccc|ccc}... & 1 & 0 \\ 0 & -1 & -1 & 1 & 0 & 0 \end{array} \right] \quad \text{by swapping } R1 \text{ and } R3\\ \
- Question 5, Exercise 2.3 @math-11-nbf:sol:unit02
- & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right]$. ** Solution. ** Given \begin{align*} A &= \... & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right]\\ |A|&= 1 [-1 - 2] + 1 [-2 + 1] + 1 [-4 - 1] \\... t|\begin{array}{cc} 1 & -1 \\ -2 & -1 \end{array}\right|\\ &= (1) [(1)(-1) - (-1)(-2)] \\ &= -1 - 2 = -3 ... ft|\begin{array}{cc} 2 & -1 \\ 1 & -1 \end{array}\right|\\ &= (-1) [(2)(-1) - (-1)(1)] \\ &= -1 [-2 + 1]
- Question 4, Exercise 2.2 @math-11-nbf:sol:unit02
- eft[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]A\left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right]&=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\end{align} ** Solution. ** Let $ B = \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right] $ and $ C = \left[\begin{array}{cc} 1 & 3 \\ 2 &
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- (i)==== For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\right|=\left|-z_{1}\right|=\left|\overline{z_{!}}\right|=\left|-\overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \
- Question 3, Exercise 1.4 @math-11-nbf:sol:unit01
- =====Question 3(i)===== If $\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \ldots\left(x_{n}+i y_{n}\right)=a+i b$, show that: (i) $\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{
- Question 7, Exercise 2.3 @math-11-nbf:sol:unit02
- =\left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right]$. ** Solution. ** Given: \begin{align*} A &= \left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &= \dfrac{1}{4} ... left[\begin{array}{ll}6 & -1 \\ -8 & 2\end{array}\right]\\ & = \left[\begin{array}{ll} \frac{6}{4} & \fra
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- }5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** Solution. ** \begin{align*}&\quad\left[ \b... 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \... 3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \beg... 2}{5} & \frac{15}{5} \\ 2 & 10 & 6 \end{array} \right]\quad R2 - 3 \cdot R1\\ \sim & \text{R}\left[ \be
- Question 7 and 8, Exercise 4.8 @math-11-nbf:sol:unit04
- e \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*} N... e \begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*} Using value of $A$ and $B$ in (1), we get \begi... &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \end{align*} Taking sum \begin{align*} S_n&=\sum
- Question 9, Exercise 8.1 @math-11-nbf:sol:unit08
- \alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1... -\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}} \\ & =\sqrt{\d... alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\ &= -\fra
- Question 5 and 6, Exercise 8.1 @math-11-nbf:sol:unit08
- \sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*... tan^2\beta} \\ &=-\sqrt{1+{{\left(-\dfrac{5}{12} \right)}^{2}}}\\ &=-\sqrt{1+\dfrac{25}{144}} \\ & =-\sqr... } =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac