Table of Contents

EGR 6013 Homework Set 2 (32-36)

— David Wagner 2009/09/19 13:38

Section 1.8.

Problem 32.

If [A] is an m × n matrix, show that each of the three elementary operations on rows of [A] can be accomplished by premultiplying [A] by a matrix [P], where [P] is formed by performing that operation on corresponding rows of the unit matrix [I] of order m. In each case, show also that [P] is nonsingular.

Problem 33.

If [A] is an m × n matrix, show that each of the three elementary operations on columns of [A] can be accomplished by postmultiplying [A] by a matrix [Q], where [Q] is formed by performing that operation on corresponding rows of the unit matrix [I] of order n. In each case, show also that [Q] is nonsingular.

Section 1.9.

√Problem 35.

(a) By investigating ranks of relevant matrices, show that the following set of equations posses a one-parameter family of solutions:

  • 2 x_1 - x_2 - x_3 = 2,
  • x_1 + 2 x_2 + x_3 = 2,
  • 4 x_1 - 7 x_2 -5 x_3 = 2.

The rank of the upper triangular matrix in the equivalent expression below is two though its order is three; it is rank-deficient by one. This indicates the remaining two variables can be expressed in terms of each other and one additional parameter.

(b) Determine the general solution.

delim{[}{ matrix{3}{3}{ 2 {-1} {-1} 1 2 1 4 {-7} {-5} } }{]} delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace} = delim{lbrace}{ matrix{3}{1}{ 2 2 2 } }{rbrace} ⇒ delim{[}{ matrix{3}{3}{ 1 {-1/2} {-1/2} 1 2 1 4 {-7} {-5} } }{]} delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace} = delim{lbrace}{ matrix{3}{1}{ 1 2 2 } }{rbrace} ⇒ delim{[}{ matrix{3}{3}{ 1 {-1/2} {-1/2} 0 {5/2} {3/2} 0 {-5} {-3} } }{]} delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace} = delim{lbrace}{ matrix{3}{1}{ 1 1 {-2} } }{rbrace} ⇒ delim{[}{ matrix{3}{3}{ 1 {-1/2} {-1/2} 0 {1} {3/5} 0 {-5} {-3} } }{]} delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace} = delim{lbrace}{ matrix{3}{1}{ 1 {2/5} {-2} } }{rbrace} ⇒ delim{[}{ matrix{3}{3}{ 1 {-1/2} {-1/2} 0 {1} {3/5} 0 0 0 } }{]} delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace} = delim{lbrace}{ matrix{3}{1}{ 1 {2/5} 0 } }{rbrace}

  • x_3 = c_3
  • x_2 + {3/5}c_3 = {2/5} ⇒ x_2 = {2/5} -{3/5}c_3
  • x_1 - {1/5} -{3/10}c_3 -{1/2}c_3 = 1 ⇒ x_1 = {6/5} -{3/10}c_3 +{1/2}c_3 = {6/5}-{1/5}c_3

Problem 36.

(a) Show that the set

  • 2 x_1 - 2 x_2 + x_3 = lambda x_1,
  • x_1 - 3 x_2 + 2 x_3 = lambda x_2,
  • -x_1 + 2 x_2 +0 x_3 = lambda x_3.

can possess a nontrivial solution only if λ = 1 or λ = -3.

(b) Obtain the general solution in each case.

delim{[}{ matrix{3}{3}{ 2 {-2} 1 1 {-3} 2 {-1} 2 0 } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = lambda delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{[}{ matrix{3}{3}{ lambda 0 0 0 lambda 0 0 0 lambda } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} ⇒ delim{[}{ matrix{3}{3}{ {2-lambda} {-2} 1 1 {-3-lambda} 2 {-1} 2 {0-lambda} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{lbrace}{matrix{3}{1}{ 0 0 0 }}{rbrace} = delim{[}{ matrix{3}{3}{ {2-lambda} {-2} 1 0 { {-3-lambda}+{{2}/{2-lambda}} } { 2-{1/{2-lambda}} } 0 {2-{2/{2-lambda}}} {-lambda}+{1/{2-lambda}} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{[}{ matrix{3}{3}{ {2-lambda} {-2} 1 0 { {-3-lambda}+{{2}/{2-lambda}} } { 2-{1/{2-lambda}} } 0 { {2-2lambda}/{2-lambda} } {{-lambda}+{1/{2-lambda}} + {1-1/{2-lambda}}} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{[}{ matrix{3}{3}{ {2-lambda} {-2} 1 0 { {(2-lambda)(-3-lambda)+2}/{2-lambda} } { {3-2lambda}/{2-lambda} } 0 { {2-2lambda}/{2-lambda} } {{-lambda}+{1/{2-lambda}} + {1-1/{2-lambda}}} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{[}{ matrix{3}{3}{ {2-lambda} {-2} 1 0 1 { {(2-lambda)(3-2lambda)}/{((2-lambda)(-3-lambda)+2)(2-lambda)}} 0 { {2-2lambda}/{2-lambda} } {{-lambda}+{1/{2-lambda}} + {1-1/{2-lambda}}} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{[}{ matrix{3}{3}{ {2-lambda} {-2} 1 0 1 { {(3-2lambda)}/{((2-lambda)(-3-lambda)+2)}} 0 0 {{-lambda}+{1/{2-lambda}} + {1-1/{2-lambda}} - {(2-2lambda)(3-2lambda)}/{((2-lambda)(-3-lambda)+2)(2-lambda)} } } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{lbrace}{matrix{3}{1}{ 0 0 0 }}{rbrace}

FIXME The following comes from a prior erroneously reduced equation.

{{-lambda x_3}+{x_3/{2-lambda}} + {x_3-x_3/{2-lambda}}} = 0 = (1-lambda)x_3 =0 ⇒ When x_3 ne 0, lambda = 1.

When x_3 = 0,

  • { {-3x_2-lambda x_2}+{{2x_2}/{2-lambda}} }= 0 = (2-lambda)(-3-lambda)x_2 + 2x_2
  • ⇒ x_2 = -{1/2}(2-lambda)(-3-lambda).
  • (2-lambda)x_1 + -2({-{1/2}}(2-lambda)(-3-lambda) ) = 0 = (2-lambda)x_1 + (2-lambda)(-3-lambda) = 0 ⇒ x_1 = 3+lambda

Going back to the last row of the very first expression,

-3-lambda -(2-lambda)(-3-lambda)=0 =-3-lambda-lambda^2 -lambda +6=-lambda^2 -2lambda +3=-(lambda+3)(lambda-1)=0

Only lambda = -3 and lambda = 1 satisfy this system for any x_3.

For lambda = 1:

delim{[}{ matrix{3}{3}{ 1 {{-2}/{2-lambda}} {1/{2-lambda}} 0 { {-3-lambda}+{{2}/{2-lambda}} } { 2-{1/{2-lambda}} } 0 0 {1-lambda} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{lbrace}{matrix{3}{1}{ 0 0 0 }}{rbrace} = delim{[}{ matrix{3}{3}{ 1 {{-2}/{2-1}} {1/{2-1}} 0 { {-3-1}+{{2}/{2-1}} } { 2-{1/{2-1}} } 0 0 0 } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{[}{ matrix{3}{3}{ 1 {-2} 1 0 {-2} 1 0 0 0 } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace}

  • x_3 = c_3 ne 0,
  • -2x_2 +c_3=0 ⇒ x_2 = {c_3/2}
  • x_1 -c_3 +c_3 = 0 ⇒ x_1 = 0

This is the general solution.

For lambda = -3:

delim{[}{ matrix{3}{3}{ 1 {{-2}/{2-lambda}} {1/{2-lambda}} 0 { {-3-lambda}+{{2}/{2-lambda}} } { 2-{1/{2-lambda}} } 0 0 {1-lambda} } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace} = delim{lbrace}{matrix{3}{1}{ 0 0 0 }}{rbrace} = delim{[}{ matrix{3}{3}{ 1 {{-2}/{5}} {1/5} 0 { 2/5 } { 9/5 } 0 0 4 } }{]} delim{lbrace}{matrix{3}{1}{ x_1 x_2 x_3 }}{rbrace}

  • x_3 = 0,
  • x_2=0,
  • x_1=0,

a trivial solution.


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