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Question 5, Exercise 1.3 @math-11-kpk:sol:unit01
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}}{2}.\end{align} Thus the solutions of the given equations are $\dfrac{1\pm\sqrt{5}}{2}$. =====Question 5(ii... t{1-i}\end{align} Thus the solutions of the given equations are $1\pm\sqrt{1-i}$. =====Question 5(iv)===== F... z=2i.\end{align} Thus, the solutions of the given equations are $\pm 2i$. ====Go To==== <text align="left"><
Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
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c{1\pm \sqrt{3}i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}... 1\pm \sqrt{3}i$ are the solutions of the required equations. =====Question 6(iii)===== Find the solutions of... i}{2}\end{align} Thus, the solutions of the given equations are $1$, $-\dfrac{1}{2}\pm \dfrac{\sqrt{3}i}{2}$.
Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
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B=0 \text { and } A-B=1 $$ Solving the above two equations for $A$ and $B$, we get \begin{align}A&=\dfrac{1}... \text{and} \quad 2 A-B=1$$ Solving the above two equations $$A=\dfrac{1}{3}\quad\text{and} B=-\dfrac{1}{3}$$... d \text{ and} \quad 4 A-5 B=1$$ Solving the above equations for $A$ and $B$ we get $A=\dfrac{1}{9}$ and $B=-
Question 5 & 6, Exercise 2.1 @math-11-kpk:sol:unit02
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}{2}.$$ =====Question 6(i)===== Solve the matrix equations for $X.$ Find $X-3A=2B$, if $A=\begin{bmatrix} 1 ... lign} =====Question 6(ii)===== Solve the matrix equations for $X.$ Find $2( X-A )=B$, if $A=\begin{bmatrix}
Question 4 Review Exercise @math-11-kpk:sol:unit05
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\cdot n$ and constant terms on the both sides of equations, we get $$A+B+C=0 \quad 15 A+6 B-3 C=0$$ $$4 A+8 B-2 C=1$$ Solving these three equations for the constants $A, B$ and $C$ we get $A=\dfra
Unit 01: Complex Numbers (Solutions) @math-11-kpk:sol
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complex numbers. * Solve simultaneous linear equations with complex coefficients. * Write the polyno
Unit 02: Matrices and Determinants (Solutions) @math-11-kpk:sol
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of complex numbers. * Solve simultaneous linear equations with complex coefficients. * Write the polynomi
Question 11 & 12 Exercise 4.5 @math-11-kpk:sol:unit04
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)^{p-q}.\end{align}\\ Multiplying the above three equations\\ \begin{align}a^{q-r} b^{r-p} c^{p-q}& =(a_1 r^{
Question 2 & 3 Exercise 5.4 @math-11-kpk:sol:unit05
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d \text{and}\quad 2 A-B=1$$ Solving the above two equations for $A$ and $B$ we get $$A=\dfrac{1}{3}\quad\tex
Question 7 and 8 Exercise 7.3 @math-11-kpk:sol:unit07
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rac{1}{4}=2-\frac{3 x^2}{16}$. From the above two equations, we get that $a \cdot b x^2=2-\frac{3 x^2}{16}$ $