Problems about Matrices, Determinants, Systems of linear equations

• READ THIS FIRST
  
• Problems about Matrices, Determinants, Systems of linear equations
  
  • Level 2 problems
    
  • Level 3 problems
    

READ THIS FIRST

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Matrices, Determinants, Systems of linear equations

Level 2 problems

• 
  

  A and B are n x n matrices. Investigate if (A + B)2 = A2 + 2.A.B + B2

  
  

   
          (A + B)2  = (A + B).(A + B) = A.A + A.B + B.A + B.B
  
                   = A2 + A.B + B.A + B2
  
  Since the product is not commutative, we do NOT have
  
          (A + B)2  = A2 + 2.A.B + B2
  

• 
  

  We say that two matrices A and B commute, if and only if AB = BA. Show that all matrices of the form [a -b] [b a] commute.

  
  

   
          [a   -b] [a'   -b']    = [aa'-bb'   -ab'-ba']
          [b    a] [b'    a']      [ba'+ab'   -bb'+aa']
  
  and
  
          [a'   -b'] [a   -b]    = [aa'-bb'   -ab'-ba']
          [b'    a'] [b    a]      [ba'+ab'   -bb'+aa']
  
  

• 
  

  Prove that for each n x n matrix A , AT .A is symmetric

  
  

   
  We'll prove that AT .A is equal to its transpose.
  
  (AT .A)T  = AT .(AT)T  = AT .A
  

• 
  

  Prove that if matrix A is skew-symmetric, then A.A is symmetric.

  
  

   
  A is skew-symmetric <=> A = - AT .
  So,
  
  (A.A)T = AT .AT = (- A ).(- A ) = A.A
  
  Hence, A.A is symmetric.
  

• 
  

  Given X = [x y] and A is a 2 x 2 matrix . All elements of the matrices are real numbers. Prove that X . AT . A . XT can't be negative.

  
  

   
  First we see that X . AT . A . XT  is a 1 x 1 matrix.
  It is a number, so maybe it can be  negative.
  But,
  
          X . AT  = is a 1 x 2 matrix, say [a   b].
  
          A . XT  = (X . AT )T  = [a   b]T
  
  Then
          X . AT . A . XT   = (X . AT ).(X . AT )T
  
                          = [a   b] .[a   b]T
  
                          = a2  + b2
  
  So,     X . AT . A . XT  can not be  negative.
  

• 
  

  Determine for which values of a and b the following system has no solutions. / ax + y + 2z = 0 | x + 2y + z = b \ 2x + y + az = 0

  
  
  The matrix formed by the coefficients is

   
  [ a  1  2]
  [ 1  2  1]
  [ 2  1  a]
  

  and the determinant D= 2a² - 2a - 4.
  D = 0 <=> a = 2 of a = -1

  1. 
     If D not 0, then the system has exactly 1 solution.
  2. 
     If a = 2, the matrix formed by the coefficients is

      
     [ 2  1  2]
     [ 1  2  1]
     [ 2  1  2]
     

     The rank of this matrix is 2. We can choose the third equation as side equation and z as a side unknown.

     The characteristic determinant is here

      
     | 2  1  0|
     | 1  2  b| = 0
     | 2  1  0|
     

     The system has always solutions when a = 2.
  3. 
     If a = -1, the matrix formed by the coefficients is

      
     [-1  1  2]
     [ 1  2  1]
     [ 2  1 -1]
     

     The rank of this matrix is 2. We can choose the third equation as side equation and z as a side unknown.

     The characteristic determinant is now

      
     |-1  1  0|
     | 1  2  b| = 3b
     | 2  1  0|
     

     If b not 0, the system has no solutions

  Conclusion: the system has no solutions if and only if (a = -1 and b not 0)

• 
  

  Calculate the m-values such that the following system has more than one solution. / (m+2)x + 2y + 4z = 3m | -mx + 5y + 2mz = -2m \ 2x + 7y + 6z = 1

  
  
  If the determinant of the matrix of the coefficients is not zero, then the system has a unique solution. So, this determinant must be zero. This is the case for m = 1 or m = -10/7.

  Case : m = 1

  If we substitute 1 for m in the system, we immediately see that the third equation is the sum of the previous equations. The third equation is superfluous. We cancel this equation. Now, the rank of the matrix of the coefficients is 2 and we can choose z as a side-unknown. The system has infinitely many solutions.

  Case : m = -10/7

  we substitute -10/7 for m in the system.We simplify a bit.

   
  / 4x + 14y + 28z = -30
  | 10x + 35y - 20z = 20
  \ 2x + 7y + 6z = 1
  

  The rank of the matrix of the coefficients is 2. We can't use x an y as main unknowns. The unknowns y and z can be used as main unknowns with the first and second equation as main equations. The system has a solution if and only if the characteristic determinant of the side equation is zero. In this system, the characteristic determinant is not zero. The system has no solutions if m = -10/7.

• 
  

  [ 1 m - 1 2 m - 3 ] Given A = [ m 2 m - 2 2 ] [ m + 1 3 m - 3 m.m - 1 ] Calculate the condition for m such that A is regular. Assume that m satisfies this condition and consider the system with A as coefficient matrix. / x + (m - 1)y + (2 m - 3)z = 1 | | m x + (2 m - 2)y + 2 z = 0 | \ (m + 1)x + (3 m - 3)y + (m.m - 1)z = 0 Now, let x = 1 and calculate the values of m such that the system has a solution for y and z.

  
  

   
                            2
  Det(A) = m (1 - m) (m - 2)
  
  A is regular if and only if  m is NOT  in { 0, 1, 2 }
  
  With x = 1, the system becomes
  
          /  (m - 1)y    + (2 m - 3)z      = 0
          |
          | (2 m - 2)y   +    2  z         =  - m
          |
          \ (3 m - 3)y   + (m.m - 1)z     = -(m + 1)
  
  The coefficient matrix of the first and the second equation is
  
          [m - 1      2 m - 3  ]
          [2 m - 2      2      ]
  

  The determinant is -4 (m - 1) (m - 2) and since m is not in { 0, 1, 2 }, the determinant is not 0.
  So, we consider the first and the second equation as main equations and the system has a solution for y and z if and only if the characteristic determinant of the third equation is 0.

   
  The condition is
          |m - 1      2 m - 3        0  |
          |2 m - 2      2          - m  | = 0
          |3 m - 3    m.m - 1    - m - 1|
  <=>
            2
          (m  + 4) (m - 1) (m - 2) = 0
  

  Since m is not in { 0, 1, 2 }, there are no real values of m such that the system has a solution for y and z.

• 
  

  The matrix x is a 2 x 2 matrix. Calculate three solutions of the quadratic equation x2 - x = 0

  
  

   
  
          x2  -  x = 0
  <=>
          x (x-I) = 0
  

  x = 0-matrix and x = Identity matrix are two solutions, but we know that x (x-I) can be zero without x = 0 or x = I.

  To find another solution of the given equation we'll search for a matrix

   
               [ 1      b ]
          x =  [          ]  such that    x2  = x
               [ c      d ]
  
          x2  = x
  <=>
          [ 1      b ] [ 1      b ]   [ 1      b ]
          [          ].[          ] = [          ]
          [ c      d ] [ c      d ]   [ c      d ]
  
  <=>
          [ 1 + b c    b + b d  ]    [ 1      b ]
          [                     ] =  [          ]
          [ c + d c   c b + d2 ]    [ c      d ]
  <=>
          [ b c            b d    ]    [0       0 ]
          [                       ] =  [          ]
          [ c d     b c + d2 - d ]    [0       0 ]
  

  Since we need only one solution, we choose d = 0 and c = 0.
  Then b = an arbitrary number. So, even with d = 0 and c = 0, we find an infinity number of solutions.

   
               [ 1      b ]
          x =  [          ]
               [ 0      0 ]
  

Level 3 problems

• 
  

  [-2 -9 ] n [1-3n -9n ] Given A = [ ] . Prove that A = [ ] [ 1 4 ] [ n 1+3n ]

  
  

   
  
  We'll prove this by complete induction.
  
  a) The property is trivial for n = 1
  
  b) Assume the property is true for n = k.
  
  Then we have to prove that the property is true for n = k+1.
  
   k+1    k     [1-3k   -9k ] [-2     -9 ]   [-2-3k  -9-9k]
  A    = A .A = [           ].[          ] = [            ]
                [ k    1+3k ] [ 1      4 ]   [k+1     3k+4]
  
          [1-3(k+1)   -9(k+1) ]
       =  [                   ]
          [ k+1      1+3(k+1) ]
  

• 
  

  [1 1 0 ] Given : A = [0 1 0 ] [0 0 1 ] Show that A is regular. n Calculate A . Calculate the real numbers a and b such that A2 + a A + b I = 0 ( I is the 3 x 3 identity matrix) Show that there are real numbers c0,c1,c2, ... ,cn such that A-n = c0.I + c1.A2 + c2.A3 + c3.A4 + c4.A + ... + cn.An

  
  

   
  
  Det(A) = 1 , so A is regular.
  
             2    [1   2   0 ]             3    [1   3   0 ]
            A =   [0   1   0 ]            A =   [0   1   0 ]
                  [0   0   1 ]                  [0   0   1 ]
  
  From this we'll prove by complete induction that
  
           n    [1   n   0 ]
          A =   [0   1   0 ]
                [0   0   1 ]
  
  This expression is true for n = 1
  Assume the property is true for n = k.
  Then we have to prove that the property is true for n = k+1.
  
  
   k+1    k     [1   k   0 ][1   1   0 ]   [ 1, k + 1, 0 ]
  A    = A .A = [0   1   0 ][0   1   0 ] = [ 0,   1,   0 ]
                [0   0   1 ][0   0   1 ]   [ 0,   0,   1 ]
  
                          A2  + a A + b I = 0
  <=>
          [1   2   0 ]     [1   1   0 ]     [1   0   0 ]   [0   0   0 ]
          [0   1   0 ] + a [0   1   0 ] + b [0   1   0 ] = [0   0   0 ]
          [0   0   1 ]     [0   0   1 ]     [0   0   1 ]   [0   0   0 ]
  
  <=>
          1 + a + b = 0
          2 + a =  0
  <=>
          a = -2, b = 1
  
  From this we have
  
  
          A2  - 2 A + I = 0
  
  <=>
  
          2 A - A2 = I
  
  <=>
  
          A (2 I - A ) = I
  
  From this, we see that (2 I - A ) is the inverse of A
  
          A-1  = 2 I - A
  
  
  =>      A-n  =  (2 I - A )n
  
  
  
  =>      A-n  = a polynomial in A of degree n, with real coefficients
  
  =>      there are real numbers  c0,c1,c2, ... ,cn such that
  
  
          A-n   = c0.I + c1.A  + c2.A2 +  c3.A3  + c4.A4  +  ...  + cn.An