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Question 4, Exercise 1.1 @math-11-nbf:sol:unit01
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2+i)}=\dfrac{(1-5i)(2-i)+y(3-2i)}{(3-2i)(2-i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2+5i^2-10i-i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2-5-11i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(3y-3)+(-11-2y)i}{(4-7i)}\\ \Rightarrow \quad & x(4-7i)=((3y-3)+(-11-2y)i)(2+i)\\ \Righta
Question 3, Exercise 2.6 @math-11-nbf:sol:unit02
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\begin{align*} \frac{19}{4}z& = -\frac{63}{4}\\ \Rightarrow z& = -\frac{63}{19}\end{align*} From the second row, we have: \begin{align*} &- 2y - 3z = 3\\ \Rightarrow &-2y - 2(-\frac{63}{19}) = 3 \\ \Rightarrow &-2y = -\frac{132}{19} \\ \Rightarrow &y = \frac{66}{19}\end{align*} Finally, from the first row, we have: \
Question 2, Exercise 6.3 @math-11-nbf:sol:unit06
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e \begin{align*}{ }^{n} C_{r}&={ }^{n} C_{n-r}\\ \Rightarrow \quad \text{If} r\quad &=7\quad\text{then}\\ n-r&=15\\ \text{or} \quad n-7&=15 \\ \Rightarrow n &=15+7=22\end{align*} =====Question 2(iii)====... w \begin{align*}{ }^{n} C_{1}&={ }^{n} C_{n-1}\\ \Rightarrow 50&=n-1\\ \Rightarrow n& =51\end{align*} =====Question 2(iv)===== Find $n$, if : $\,\, ^{2n}C_3:^nC_3=\,\,1
Question 3, Exercise 1.2 @math-11-nbf:sol:unit01
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n we $z$ is real. As \begin{align}& z=\bar{z}\\ \Rightarrow \quad & a+ib=a-ib\\ \Rightarrow \quad & 2ib=0\\ \Rightarrow \quad & b=0\quad \because \quad 2i\neq 0\end{align} Then (1) becomes $$z=a+i(0)=a
Question 4, Exercise 1.2 @math-11-nbf:sol:unit01
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\end{align} Now \begin{align}&|z_{1} z_{2}|=16\\ \Rightarrow \quad &|z_{1}|| z_{2}|=16\\ \Rightarrow \quad & \sqrt{13}|z_2|=16\\ \Rightarrow \quad & |z_2|=\dfrac{16}{\sqrt{13}}\end{align} GOOD ====Go to ==== <
Question 4, Exercise 2.6 @math-11-nbf:sol:unit02
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\frac{3}{2}x_2 + \frac{7}{2}x_3 = \frac{1}{2}\\ \Rightarrow &x_1 = \frac{1}{2} + \frac{3}{2}\frac{2}{11}\\ \Rightarrow & x_1= \frac{1}{2} + \frac{3}{11}\\ \Rightarrow & x_1= \frac{17}{22}\end{align*} Thus, the general solution i
Question 1(i-v), Exercise 6.3 @math-11-nbf:sol:unit06
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\\ $$ \begin{array}{ll} & r!X={ }^{n} P_{r} \\ \Rightarrow & r!X=\frac{n!}{(n-r)!} \\ \Rightarrow & X=\frac{n!}{r!(n-r)!}={ }^{n}{C}_{r} \end{array} $$ =====Question 1(... r)!} \\ & \quad \because(n-r)!(n-r+1)=(n-r+1)!\\ \Rightarrow \quad&=(n-r+1) \frac{n!}{(r-1)!(n-r+1)!}\\ & =(n-
Question 5 and 6, Exercise 8.1 @math-11-nbf:sol:unit08
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=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*} Als... } =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac{12}{... }} =-\dfrac{17}{8} \end{align*} \begin{align*} \Rightarrow \quad \sin \beta=\frac{1}{\csc \beta}=-\frac{8}{1
Question 1(vi-x), Exercise 6.3 @math-11-nbf:sol:unit06
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Let \begin{align*}{ }^{n} C_{p}&={ }^{n} C_{q}\\ \Rightarrow \dfrac{n!}{p!(n-p)!}&=\dfrac{n!}{q!(n-q)!}\\ q!(n-q)!&=p!(n-p)!\\ \Rightarrow q!&=p!\quad i.e. \quad q=p \quad \\ \text{or} \qu
Question 1, Exercise 6.2 @math-11-nbf:sol:unit06
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}{(n-n+1)!}\\ &=\dfrac{n!}{1!}=n!\because 1!=1\\ \Rightarrow\quad L.H.S &= R.H.S\end{align*} =====Question 1(
Question 2, Exercise 6.2 @math-11-nbf:sol:unit06
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, \\ & n=8 \end{align*} $n$ must be an integer, $\Rightarrow n=8$ =====Question 2(ix)===== Find $n$, if: $\qu
Exercise 6.3 (Solutions) @math-11-nbf:sol:unit06
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2 n-1)]}{n!}$ (vii) ${ }^{n} C_{p}={ }^{n} C_{q} \Rightarrow p=q$ or $p+q=n$ \\ (viii) ${ }^{n} C_{r}+2{ }^{n}