Math 310: Hour Exam 2 (Solutions)

Prof. S. Smith: Fri 4 Nov 1994

You must SHOW WORK to receive credit. Problem 1:

(a) Are the vectors ( \begin{matrix}2 \\ −1 \\ 0 \end{matrix} ), ( \begin{matrix}−1 \\ 2 \\ −1 \end{matrix} ), ( \begin{matrix}0 \\ −1 \\ 2 \end{matrix} ) linearly independent?

YES: Maybe easiest is det(A) = 4 ≠ 0. Or check rref(A)= identity .

(b) Do the vectors ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ), ( \begin{matrix}4 \\ 5 \\ 6 \end{matrix} ), ( \begin{matrix}7 \\ 8 \\ 9 \end{matrix} ) form a spanning set for R^{3}?

NO: find det(A) = 0; or compute rref(A) as ( \begin{matrix}1 & 0 & −1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix} )—no pivot in 3rd column.

Problem 2:

(a) Find a basis for the row space of ( \begin{matrix}1 & 3 & 3 & 2 \\ 2 & 6 & 9 & 5 \\ −1 & −3 & 3 & 0 \end{matrix} ) . Is (1, 2, 3, 0) in this row space?

Compute rref(A) as ( \begin{matrix}1 & 3 & 0 & 1 \\ 0 & 0 & 1 & \frac{1}{3} \\ 0 & 0 & 0 & 0 \end{matrix} ) ; first two rows give a basis.

And (1, 2, 3, 0) is NOT in the space—because of pivots in 1st and 3rd columns,

it would have to equal the combination 1(1, 3, 0, 1) + 3(0, 0, 1, \frac{1}{3}); but doesn’t.

(b) Find a basis for the subspace of all symmetric (A = A^{T}) 2x2 matrices. What is the dimension of this subspace?

By the symmetric requirement, these matrices have form ( \begin{matrix}a & b \\ b & d \end{matrix} ) .

Choosing a = 1, b = 0, d = 0 etc, get matrices

( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) , ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) for basis. And dimension is 3.

Problem 3: Find the matrix, in the standard bases, for the linear transformation

L(( \begin{matrix}x \\ y \\ z \end{matrix} )) = ( \begin{matrix}2z \\ 3x + y \\ 2x − z \end{matrix} ) .

Apply L to standard basis, put into columns to get ( \begin{matrix}0 & 0 & 2 \\ 3 & 1 & 0 \\ 2 & 0 & −1 \end{matrix} ) .

Problem 4:

(a) Find an orthonormal basis for the subspace W of R^{3} spanned by the row vectors (1, 1, 1) and (1, 1, 0).

Let’s use Gram-Schmidt, with normalization afterwards. Call initial vectors x_{1},x_{2}. Form y_{2} = x_{2} − \frac{x_{2}·x_{1}}{x_{1}·x_{1}} x_{1} = (1, 1, 0) − \frac{2}{3}(1, 1, 1) = \frac{1}{3}(1, 1, −2). Now normalize to get u_{1} = \frac{1}{\sqrt{3}}(1, 1, 1) and u_{2} = \frac{1}{\sqrt{6}}(1, 1, −2).

Note: using (1, 1, 0) first leads to nicer basis \frac{1}{\sqrt{2}}(1, 1, 0) and (0, 0, 1).

(b) Now find a basis for the subspace W^{⊥} orthogonal to W in part (a).

Using the vectors in (a) as rows of A and row-reducing gives ( \begin{matrix}1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ).

Nullspace has form a(1, −1, 0); normalizing gives basis \frac{1}{\sqrt{2}}(1, −1, 0).

Problem 5: Find the least-squares solution of the inconsistent system

( \begin{matrix}1 & 0 \\ 1 & 3 \\ 1 & 6 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} ) = ( \begin{matrix}1 \\ 4 \\ 5 \end{matrix} ) .

What is the error with this approximation?

Multiply A^{T} by the augmented matrix [Ab] to get

( \begin{matrix}1 & 1 & 1 \\ 0 & 3 & 6 \end{matrix} ) ( \begin{matrix}1 & 0 & 1 \\ 1 & 3 & 4 \\ 1 & 6 & 5 \end{matrix} ) = ( \begin{matrix}3 & 9 & 10 \\ 9 & 45 & 42 \end{matrix} ) .

Row operation A_{2}^{−3×1} leads to ( \begin{matrix}3 & 9 & 10 \\ 0 & 18 & 12 \end{matrix} ). This gives y = \frac{2}{3} and then x = \frac{4}{3}.

Multiply by A to get approximation \frac{1}{3} ( \begin{matrix}4 \\ 10 \\ 16 \end{matrix} ), so error is \frac{1}{3} ( \begin{matrix}−1 \\ 2 \\ −1 \end{matrix} ), of length \sqrt{\frac{2}{3}}.