MATH 2050 Linear Algebra (Section 4)

Department of Mathematics and Statistics
2009 Winter

Term Test 2 - Solutions

2009 March 13

  1. Find the value of the determinant of each of the following matrices.
    1.   A = [ 3 0 0 0 ] [ x -2 0 0 ] [ 17 y 1 0 ] [ pi 42 -6 4 ]

      A   is a lower triangular matrix.
      Its determinant is therefore the product of its diagonal elements:   3 × (–2) × 1 × 4     Þ

      det A = -24

    2.   B = (a 5x5 matrix with rows 2 & 4 the same)

      Rows 2 and 4 are identical.   Therefore, without any further calculation,

      det B = 0

    3.   C = [ 4 -2 1 5 ] [ 2 9 0 3 ] [ 0 2 0 0 ] [ 5 8 0 7 ]

      Cofactor expansion down column 3:   det C = | 2 9 3 | | 0 2 0 | | 5 8 7 |
      Cofactor expansion along row 2:   det C = 2 × | 2 3 | | 5 7 |

      det C = -2

  1. Find all eigenvalues and corresponding eigenvectors of the matrix
              A = [ 7 0 -4 ] [ 0 5 0 ] [ 5 0 -2 ]
    Hence write down both the matrix   P   that diagonalizes   A   and the diagonal matrix   D = P^-1 AP.
    [Note:   you do not have to evaluate the matrix product   P^-1 AP.]

    The characteristic equation for matrix   A   is
    c = det(lambda*I - A) = 0
    Conducting a cofactor expansion along row 2 or down column 2:
    c = ...
    c = (L-5)(L-3)(L-2) = 0

    lambda = 2 or 3 or 5

    [Note that the fact that all three eigenvalues are distinct (and are therefore all of multiplicity 1) guarantees that the matrix   A   is diagonalizable.]

    For eigenvalue   l = 2:

    (2I-A)X = O ...
    -5x + 4z = 0 and y = 0
    Therefore the set of (2)-eigenvectors of   A   is all non-zero multiples of the basic eigenvector
                X2 = [ 4 0 5 ]T

    For eigenvalue   l = 3:

    (3I-A)X = O ...
    -x + z = 0 and y = 0
    Therefore the set of (3)-eigenvectors of   A   is all non-zero multiples of the basic eigenvector
                X3 = [ 1 0 1 ]T

    For eigenvalue   l = 5:

    (5I-A)X = O ...
    x = 0 and z = 0
    Therefore the set of (5)-eigenvectors of   A   is all non-zero multiples of the basic eigenvector
                X5 = [ 0 1 0 ]T

    Hence the matrices   P   and   D   are

    P = and D = diag(2,3,5) [ 4 1 0 ] [ 0 0 1 ] [ 5 1 0 ]

    Interchanging the order of the eigenvalues in   D   will cause a corresponding interchange in the order of the columns of   P .
    Any column of   P   may be multiplied by any non-zero constant.   The inverse of   P   will be affected in such a way that the diagonal matrix   D = P^-1 AP   will not change.

  1. Evaluate   det([2A]^(-1) [3A]T), where   A   is any invertible (2×2) matrix.

    Employing the general results   det(AB) = (det A)(det B)   (for all square matrices for which the product is defined),   det(kA) = k^n (det A)   and   det(AT) = det A   (for all (n×n) matrices   A ),   and   det(A^-1) = 1 / det A   (for all invertible (n×n) matrices   A ):
    det([2A]^(-1) [3A]T) = ...
    Therefore

    det([2A]^(-1) [3A]T) = 9/4 = 2.25

      BONUS QUESTION:

  1. Find   (3,4) element of T-inverse   (the entry in row 3, column 4 of the inverse matrix   (3,4) T-inverse)
    for the matrix   T = [ 1 2 3 4 ] [ 0 5 6 7 ] [ 0 0 2 8 ] [ 0 0 0 1 ].

    inverse = adjugate / determinant,   where   C(T)   is the matrix of cofactors of   T.
    The (i, j) cofactor of   T   is   c_ij = (-1)^(i+j) det T_ij, where   Tij   is the matrix formed by deleting row i and column j from the matrix   T.   Therefore
    (3,4) element of T-inverse = -4
    [Note that it is not necessary to evaluate the entire inverse matrix of   T !]

    (3,4) element of T-inverse = -4

    [After considerable additional effort, either by Gaussian elimination or by the adjugate / determinant method, the complete inverse matrix of   T   can be shown to be
    full inverse matrix of T   .]

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    Created 2009 03 05 and most recently modified 2009 03 06 by Dr. G.H. George