Linear Algebra Study Notes
LINEAR ALGEBRA NOTES
- Course Code: 015403525
- Course Title: EL 18
- Institution: University of Mines and Technology
- Faculty: Computing and Mathematical Sciences
- Department: Mathematical Sciences
- Course Number: EL 171
Problem Set
Problem 1: Solve for z
- Given: z=−2+(2ext√3)i
- Required Form: Express the roots in the form: r(extcosheta+iextsinheta)
- Solution Steps:
- Identify the modulus, r:
- r=ext√((−2)2+(2ext√3)2)
- r=ext√(4+12)
- r=ext√(16)=4
- Identify the argument, heta:
- heta=exttan−1−22ext√3
- Since the point is in the second quadrant, adjust:
- heta=extπ−exttan−1(ext√3)=extπ−3extπ=32extπ
- Final expression:
- z=4(extcos32extπ+iextsin32extπ)
Problem 2: De-Moivre's Theorem Application
- Task: Show that an(40)=extexactrootsoftheequation
4ext√3−6−4ext√3+1=0 - Solution Steps:
- Simplifying the expression given:
- Combining like terms: 4ext√3−4ext√3−6+1=−5
- Equation simplifies to:
- −5=0 which leads to an inconsistency.
- Investigate values of tangent at related angles:
- Utilizing De-Moivre's theorem:
- ext(cosheta+iextsinheta)n=extcos(nheta)+iextsin(nheta)
- For 40exto:
- Consider multiples of tan(40exto) related to cyclical patterns in trig functions.
Problem 3: Binomial Expansion
- Given: (1+x)n=P<em>0+P</em>1x+P<em>2x2+P</em>3x3+…
- Prove that:
- an(0)−4=41−6an2(0)
- i. P=P<em>2+P</em>1=22(extcos(17t))
- ii. R=P<em>3+P</em>t=422(extsin(t))
- Implications of the Expansion:
- Relate the binomial coefficients to combinatorial interpretations, such as selection of elements from a set.