Question 1, Exercise 1.3

Solutions of Question 1 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Solve the simultaneous linear equation with complex coefficient. $$\begin{align}&z-4w=3i\\ &2z+3w=11-5i\end{align}$$

Given that $$\begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{align}$$ Multiply $2$ by (i), we get
$$\begin{align}2z-8w&=6i …(iii)\end{align}$$ Subtract (iii) from (ii), we get
$$\begin{array}{cccc} 2z&-8w&=6i \\ \mathop+\limits_{-}2z&\mathop+\limits_{-}3w&=\mathop-\limits_{+}5i&\mathop+\limits_{-}11 \\ \hline 0&-11w&=11i &-11\\ \end{array}$$ $$\begin{align}-11w&=11i-11\\ \implies w&=\dfrac{11-11i}{11}\\ \implies w&=1-i\end{align}$$ Put value of $w$ in (i). $$\begin{align}z&-4(1-i)=3i\\ \implies z &=4(1-i)+3i\\ &=4-4i+3i\\ &=4-i\end{align}$$ Hence $$z=4-i, \quad w=1-i.$$

Solve the simultaneous linear equation with complex coefficient. $$\begin{align}&z+w=3i\\ &2z+3w=2\end{align}$$

Given that $$\begin{align}z+w&=3i …(i)\\ 2z+3w&=2 …(ii)\end{align}$$ Multiply $2$ by (i), we get
$$\begin{align}2z+2w&=6i …(iii)\end{align}$$ Subtract (iii) from (ii), we get
$$\begin{array}{ccc} 2z & +2w & =6i\\ \mathop+\limits_{-}2z & \mathop+\limits_{-}3w & =\mathop+\limits_{-}2 \\ \hline -w & =6i & -2\\ \end{array}$$ $$-w=6i-2$$ $$w=2-6i$$ Put value of $w$ in (i).
$$\begin{align}z&+(2-6i)=3i\\ \implies z&=3i-2+6i\\ &=-2+9i\end{align}$$ Hence $$z=-2+9i, \quad w=2-6i.$$

Solve the simultaneous linear equation with complex coefficient. $$\begin{align} &3z+(2+i)w=11-i\\ &(2-i)z-w=-1+i\end{align}$$

Given that $$\begin{align}3z+\left( 2+i \right)w&=11-i …(i)\\ \left( 2-i \right)z-w&=-1+i …(ii)\end{align}$$ Multiply $\left( 2-i \right)$ by (ii), we get,
$$\begin{align}\left( 2-i \right)\left( 2+i \right)z-\left( 2+i \right)w&=\left( 2+i \right)\left( i-1 \right)\\ 5z-\left( 2+i \right)w&=i-3 …(iii)\end{align}$$ Add (i) and (iii), we get,
$$\begin{array}{cccc} 3z & +(2+i)w & =-i+11\\ 5z & -(2+i)w & =i-3 \\ \hline 8z & 0 & =8\\ \end{array}\\ \implies z=1$$ Put value of $z$ in (i).
$$\begin{align} &3(1)+(2+i)w=11-i\\ \implies &(2+i)w=11-i-3\\ \implies &(2+i)w=8-i\end{align}$$ $$\begin{align} \implies w&=\dfrac{8-i}{2+i}\\ &=\dfrac{8-i}{2+i}\times \dfrac{2-i}{2-i}\\ &=\dfrac{16-8i-2i-1}{4+1}\\ &=\dfrac{15-10i}{5}=3-2i\end{align}$$ Hence $$z=1, \quad w=3-2i.$$

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