Linear Algebra Study Notes

LINEAR ALGEBRA NOTES

Course Information

  • 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+(2ext3)iz = -2 + (2 ext{√}3)i
  • Required Form: Express the roots in the form: r(extcosheta+iextsinheta)r( ext{cos} heta + i ext{sin} heta)
  • Solution Steps:
    • Identify the modulus, rr:
      • r=ext((2)2+(2ext3)2)r = ext{√}((-2)^2 + (2 ext{√}3)^2)
      • r=ext(4+12)r = ext{√}(4 + 12)
      • r=ext(16)=4r = ext{√}(16) = 4
    • Identify the argument, hetaheta:
      • heta=exttan12ext32heta = ext{tan}^{-1}\frac{2 ext{√}3}{-2}
      • Since the point is in the second quadrant, adjust:
      • heta=extπexttan1(ext3)=extπextπ3=2extπ3heta = ext{π} - ext{tan}^{-1}( ext{√}3) = ext{π} - \frac{ ext{π}}{3} = \frac{2 ext{π}}{3}
    • Final expression:
      • z=4(extcos2extπ3+iextsin2extπ3)z = 4 \bigg( ext{cos}\frac{2 ext{π}}{3} + i ext{sin}\frac{2 ext{π}}{3}\bigg)
Problem 2: De-Moivre's Theorem Application
  • Task: Show that an(40)=extexactrootsoftheequationan(40) = ext{exact roots of the equation}
    4ext364ext3+1=04 ext{√}3 - 6 - 4 ext{√}3 + 1 = 0
  • Solution Steps:
    • Simplifying the expression given:
      • Combining like terms: 4ext34ext36+1=54 ext{√}3 - 4 ext{√}3 - 6 + 1 = -5
      • Equation simplifies to:
      • 5=0-5 = 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)ext{(cos} heta + i ext{sin} heta)^n = ext{cos}(n heta) + i ext{sin}(n heta)
      • For 40exto40^ ext{o}:
        • Consider multiples of tan(40exto)tan(40^ ext{o}) 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+(1+x)^n = P<em>0 + P</em>1 x + P<em>2 x^2 + P</em>3 x^3 + …
  • Prove that:
    • an(0)4=16an2(0)4an(0)-4 = \frac{1-6 an^2(0)}{4}
    • i. P=P<em>2+P</em>1=22(extcos(17t))P = P<em>2 + P</em>1 = 2^2 ( ext{cos}(17t))
    • ii. R=P<em>3+P</em>t=22(extsin(t))4R = P<em>3 + P</em>t = \frac{2^2 ( ext{sin}(t))}{4}
  • Implications of the Expansion:
    • Relate the binomial coefficients to combinatorial interpretations, such as selection of elements from a set.