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    Vectors, matrices, and polynomials

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    5) Use cofactor expansion to find the determinant
    | 4 -1 1 6 |
    | 0 0 -3 3 |
    | 4 1 0 14|
    | 4 1 3 2 |

    6)Find the characteristic equation, eigenvalues, and corresponding eigenvectors for

    A = 1 2
    2 1

    7) Answer True or False for the following statements: (True means always true; False means sometimes false). Justify your answers. Remember, you can use counterexamples only to justify False)

    a) Statement: "If T: R^n -> R^n is a one-to-one linear operator, then T is invertible".

    b) Let Ta: R^n -> R^n be a linear operator with standard matrix [Ta] = A.
    Statement: "If det (A) = 0 then the range of Ta = R^n. ".

    c) Consider the following linear operations on R²:
    T1 is a counterclockwise rotation of 60º, and T2 is an orthogonal projection on the y-axis is R².
    Statement: "T1ºT2 = T2ºT1"

    d) Statement: "T(x, y, z) = (2x, 0, 1) is a linear operator in R³."

    8) Answer True or False for the following statements: (True means always true; False means sometimes false). Justify your answers. Remember, you can use counterexamples only to justify False)

    a) The set of all invertible 3x3 matrices is a subspace in the vector space of all 3x3 matrices under standard operations of matrix addition and scalar multiplication.

    b) The set of all polynomials a0 + a1x + a2x² + a3x³ for which a0= 0 is a subspace in the vector space R3[X] of all polynomials of degree ≤3 under standard operations of polynomial addition and scalar multiplication.

    c) The set of all polynomials a0 + a1x + a2x^2 + a3x^3 for which a3 ≠ 0 is a subspace in the vector space R3[X] of all polynomials of degree ≤3 under standard operations of polynomial addition and scalar multiplication

    9)Determine whether w is in span of the given vectors v1, v2,.. , and if it is, express w as a linear combination of v1, v2, ...

    a) w = {-1, 4, 15}, v1 = ( 1,2,8) v2 = (3,0,1)

    b) w= 4x+6x², v1 = 1 + x + x², v2 = 2 + 2x + 3x³, v3 = 1+5x+6x²

    c) w= 0 5 v1= 1 2 v2= 3 1
    -4 -1 -1 0 1 1

    10) Determine whether the given vectors span R³.

    a) (1,3,4), (2,0,1), (2,1,0)

    b) (1,5,1), (2,6,1), (3,3,0), (4,6,2)

    © BrainMass Inc. brainmass.com October 9, 2019, 6:33 pm ad1c9bdddf
    https://brainmass.com/math/matrices/vectors-matrices-polynomials-91345

    Solution Preview

    (Qu. 5)
    The cofactor expansion for the determinant of a matrix, A, will give the same result no

    matter which row or column of the matrix we choose - so we can choose the row or column

    that will give us least work - in this case, row 2 looks the most likely because of the

    two zeros.

    The expansion along row i (=2) is done by summing the n cofactors. Each cofactor is the

    product of three numbers:

    1) the a[i,j] entry in the matrix,
    2) the determinant of the smaller (n-1 by n-1) matrix obtained from A by deleting row i

    and column j (we'll call the determinant of this smaller matrix: A[i,j])
    3) and a +1 or -1 factor which is calculated as (-1)^(i+j)

    In full:

    n
    det A = Sum (-1)^(i+j) a[i,j] det A[i,j]
    j=1

    So,

    | 4 -1 1 6 |
    | 0 0 -3 3 | = (-1)(0)(A[2,1]) + (1)(0)(A[2,2]) +
    | 4 1 0 14 |
    | 4 1 3 2 | (-1)(-3)(A[2,3]) + (1)(3)(A[2,4])

    = 3 A[2,3] + 3 A[2,4]

    So we need to work out A[2,3] (= B) and A[2,4] (= C):

    | 4 -1 6 |
    B = | 4 1 14 | = (1)(4)B[1,1] + (-1)(-1)B[1,2] + (1)(6)B[1,3]
    | 4 1 2 |

    to find the value of B we can again use the cofactor expansion - here we have expanded B

    along the first row.

    B[1,1] = | 1 14 | = (1)(1)(2) + (-1)(14)(1) = -12
    | 1 2 |

    We can continue carrying out the expansion until we reach 1 by 1 matrices, which have a

    determinant simply equal to the value of their single entry.

    Doing this repeatedly gives:

    B = (4)(-12) + (-48) + (6)(0)
    = -96

    Similarly, you can show that C = A[2,4] = 24

    So,

    det A = 3(-96) + 3(24) = -216

    (Qu. 6)

    Recalling the definitions, we say that a nonzero vector, v, is an eigenvector with

    corresponding eigenvalue, lambda, of matrix A, if and only if they satisfy the equation:

    A.v = lambda.v

    If you'll let me say lamdba = r (just for ease of notation, lamdba is the usual notation

    for the eigenvalue though!), we can rewrite A.v = r.v as:

    (A - rI).v = 0

    where I is the identity matrix. We want to find the values of r for which the matrix:
    (A - rI) is singular (i.e non invertible), which means the determinant of (A - rI) must

    be zero:

    | A - rI | = 0

    This is our characteristic equation.
    =>
    | 1-r 2 | = 0
    | 2 1-r |

    => (1-r)(1-r) - 4 = 0

    => r^2 - 2r - 3 = 0 factorising this gives...

    => ...

    Solution Summary

    These problems deal with cofactor expansion, eigenvalues, eigenvectors, linear operators, polynomial addition, spans, and scalar multiplication.

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