M340L Second midterm exam solutions, March 7, 2003

1. Consider the tex2html_wrap_inline53 matrix

displaymath55

a) What is the rank of A? [Note: the row reduction of A can be done using integers only. If you start getting fractions, you probably made a mistake]

After row reduction, A becomes tex2html_wrap_inline63 . There are 3 pivots, so the rank is three.

b) Find a basis for the column space of A.

The pivots are in the 1st, 2nd and 4th columns, so the 1st, 2nd and 4th columns of the ORIGINAL matrix form a basis for Col(A): tex2html_wrap_inline69

c) Find a basis for the null space of A.

displaymath73

2. Consider the linear transformation tex2html_wrap_inline75 defined by
tex2html_wrap_inline77 .

a) Find the matrix of this linear transformation.

The matrix is tex2html_wrap_inline79 , whose determinant is 2.

b) Is T one-to-one? Is T onto?

Since the determinant is nonzero, the matrix is invertible, and T is both 1-1 and onto.

c) Let S be a rectangle whose area is 2. What is the area of T(S)?

The determinant is the expansion factor for area (or, in higher dimensions, volume), so the area of tex2html_wrap_inline91 .

3. Compute the following determinants:

a) tex2html_wrap_inline93

b) tex2html_wrap_inline95

c) tex2html_wrap_inline97

4. Consider the matrix tex2html_wrap_inline99 , that factorizes as A = LU, where tex2html_wrap_inline103 and tex2html_wrap_inline105 . Let tex2html_wrap_inline107 . Solve the problem tex2html_wrap_inline109 in two steps (Note: you do NOT get any credit for solving the full equations directly):

a) Solve tex2html_wrap_inline111 for tex2html_wrap_inline113 , and then

By row reducing the augmented matrix tex2html_wrap_inline115 we get tex2html_wrap_inline117 .

b) Solve tex2html_wrap_inline119 for tex2html_wrap_inline121 . Does this solution also satisfy tex2html_wrap_inline109 ?

By row reducing tex2html_wrap_inline125 we get tex2html_wrap_inline127 . Multiplying out Ax does indeed give tex2html_wrap_inline131 . You can see this either by explicit multiplication, or since A x = LUx = L(Ux) = Ly = b.

5. Indicate whether each of these statements is true or false. If a statement is sometimes true and sometimes false, write ``false''. You do NOT have to justify your answers. There is no penalty for wrong answers, so go ahead and guess if you are unsure of your answer.

a) The rank of a matrix is the dimension of its null space.

False. The rank is the dimension of the COLUMN space.

b) Swapping two rows of a matrix does not change the determinant of that matrix.

False. Swapping two rows changes the sign of the determinant.

c) If A and B are square matrices of the same size, then tex2html_wrap_inline139 .

True.

d) If an tex2html_wrap_inline141 matrix has an inverse, then the columns of that matrix span tex2html_wrap_inline143 .

True.

e) If the determinant of a (square) matrix is zero, then its columns are linearly dependent.

True. Conversely, if the determinant is nonzero, then the columns are linearly independent.

f) The span of 3 vectors in tex2html_wrap_inline143 is a 3-dimensional subspace of tex2html_wrap_inline143 .

False. The three vectors may be linearly dependent, in which case their span would have dimension less than 3.

g) Let A be an tex2html_wrap_inline151 matrix. The dimension of Col(A) plus the dimension of Null(A) equals 5.

False. The dimension of Col(A) plus the dimension of Null(A) equals 7.

h) If two vectors tex2html_wrap_inline161 and tex2html_wrap_inline163 lie in a subspace H of tex2html_wrap_inline143 , then every linear combination of tex2html_wrap_inline161 and tex2html_wrap_inline163 also lies in H.

True.

i) If a tex2html_wrap_inline175 matrix has rank 3, then its null space is 1-dimensional.

True.

j) A basis for a subspace H is a linearly independent set of vectors whose span is H.

True.



Lorenzo Sadun
Fri Mar 7 15:14:15 CST 2003