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- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- in 180 \sin 60 \\ \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{a... in 180 \sin 60 \\ \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{al
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}... \sin 2\theta & = 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\thet... ign*} \cos 2\theta & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25}
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- on 9(iii)==== Find real and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\] \[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}
- Question 1, Exercise 2.5 @math-11-nbf:sol:unit02
- nto echelon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 &... ight]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{ccc}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 ... ad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- x-1+iy) = -\dfrac{\pi}{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-... s and inequations in Cartesian form: $z \bar{z}=4\left|e^{i \theta}\right|$ ** Solution. ** Suppose $z... en $\bar{z}=x-iy$. As \begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| \\ \implies & (x+iy)(x-iy)=4|
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- row operations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 ... } A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 ... non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0 & -1 & -1 & 1 & 0 & 0 \... & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & -1 & 4 & 0 & 0 & 1 \\
- Question 5, Exercise 2.3 @math-11-nbf:sol:unit02
- llowing matrices if it exists by adjoint method $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 &... ]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 ... x for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{array}{cc} 1 & -1 \\ -2 & -1 \end{array}\r... )(-2)] \\ &= -1 - 2 = -3 \\ A_{12} &= (-1)^{1+2} \left|\begin{array}{cc} 2 & -1 \\ 1 & -1 \end{array}\ri
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- Question 10(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{
- Question 4, Exercise 2.2 @math-11-nbf:sol:unit02
- =====Question 4(i)===== Find $A$ if \begin{align}\left[\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}\righ
- Question 3, Exercise 1.4 @math-11-nbf:sol:unit01
- lamabad, Pakistan. =====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
- Question 7, Exercise 2.3 @math-11-nbf:sol:unit02
- === Verify that $(A B)^{-1}=B^{-1} A^{-1}$ if $A=\left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]$ 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}\righ
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- n 2(i)===== Find the rank of each of the matrix $\left[\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & ... y}\right]$ ** Solution. ** \begin{align*}&\quad\left[ \begin{array}{ccc} 5 & 9 & 3 \\ 3 & -5 & 6 \\ ... 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5}... ay} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5}
- Question 7 and 8, Exercise 4.8 @math-11-nbf:sol:unit04
- dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = ... rac{1}{3}$ in (2), we have \begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*} Using value of $A$ a... ign*} This gives \begin{align*} T_k &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \end{align*}
- Question 9, Exercise 8.1 @math-11-nbf:sol:unit08
- s\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\ &=-\sqrt{1-\d... cos \beta & =\sqrt{1-\cos^2\beta} \\ &=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9... \alpha \cos \beta + \cos \alpha \sin \beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \fr
- Question 2, Exercise 1.4 @math-11-nbf:sol:unit01
- ====Question 2(i)===== Write the complex number $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$... rac{\pi}{6}} \cdot e^{i\frac{\pi}{3}} \\ & = e^{i\left(\frac{\pi}{6}+\frac{\pi}{3}\right)} \\ & = e^{i\f... = 0 + i (1) = i. \end{align} Hence, we proved $$ \left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\