# MATH 2270-1 PRACTICE EXAM 2-SP12

MATH 2270-1
PRACTICE EXAM 2-SP12
1. A must be square so that m = n and the rank of A must equal n.
2. (a) Rk(A) = n, (b) N(A) = {0}, (c) COL(A) = Rn , ROW(A) = Rn , (d)
RREF(A) = In , (e) AX = Y has a solution for any Y ∈ Rn , (f) AX = Y
has a unique solution for any Y ∈ Rn .
3. T must be one-to one and onto.
4. Let Eij denote the 2 &times; 3 matrix which has a zero in every row and column
except for a one in the i-th row and j-th column. For two examples of linear
transformations let T (E11 ) = 1, T (E12 ) = x, T (E13 ) = x2 , and T (E2j ) = 0
for j = 1, 2, 3. And let S be similarly defined except that S(E1j ) = 0 for j =
1, 2, 3.For two nonlinear transformations let T (A) = x for all A ∈ M2&times;3 , and
S(A) = 1 + a11 + a12 x + a13 x2 .


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&middot;&middot;&middot;
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5. [T ]BV ,BW =  [T (v1 )]BW &middot; &middot; &middot; [T (vn )]BW  .
|
&middot;&middot;&middot;
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6. [T (X)]BW = AX so T (X) = y1 w1 + &middot; &middot; &middot; + ym wm where Y = [T (X)]BW
7. A basis for COL(A) is obtained by taking the vectors in A that correspond
to pivot columns in RREF(A). The nonzero rows of RREF(A) give a basis for
ROW(A). To find a basis for N(A) solve RREF(A) = 0 and letting the free
variables be t1 , . . . , tp then X ∈N(A) can be written as a linear combination of
the form t1 X1 + &middot; &middot; &middot; + tp Xp and X1 , . . . , Xp gives a basis for N(A). For a basis
for N(AT ) take the system of equations Y &middot; Y
&middot; , k for all Yi in the
i = 0, i = 1, &middot; &middot;
| ...
|
basis for COL(A). Row reduce the matrix  Y1 . . . Yk  and do what was
| &middot;&middot;&middot;
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done for N(A).
8. As Rk(A) = k then BROW and BCOL must have k elements in them. Taking
BROW ∪ BN(A) for a basis for Rn and BCOL ∪ BN(AT ) as a basis for Rm we get
the desired matrix for A.
9. As BCOL ∪ BN(AT ) is a basis for Rm then Y = r1 Y1 + &middot; &middot; &middot; + rk Yk + rk+1 Yk+1 +
&middot; &middot; &middot; + rm Ym where Y1 , . . . , Yk is a basis for COL(A) and Yk+1 , . . . , Ym is a basis
for N(AT ). Then YC = r1 Y1 + &middot; &middot; &middot; + rk Yk , and YN = rk+1 Yk+1 + &middot; &middot; &middot; + rm Ym .
10. Augment the matrix AT A by the vector Y and then row reduce to get the
complete solution to the least squares problem. As k &lt; m &lt; n then there will
be n − k free variables and hence infinitely many solutions.