18.06 Exam 2 #1 Solutions

1. The row echelon form of A is [ \begin{matrix}1 & 2 & −1 & 4 \\ 0 & 1 & −2 & 3 \\ 0 & 0 & 0 & 0 \end{matrix} ]. So we find that a basis for R(A^{T}) is

{(1, 2, −1, 4), (0, 1, −2, 3)}, and a basis for N(A) is {(−3, 2, 1, 0), (2, −3, 0, 1)}. Similarly, row

echelon form of A^{T} is [ \begin{matrix}1 & 0 & −1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ]. So a basis for C(A) is \{(1, 0, −1), (0, 1, 2)\} and a basis

for N(A^{T}) is {(1, −2, 1)}.

2. a) Using the row operation R4 − R1 gives

| \begin{matrix}−1 & 2 & 0 & 1 \\ 1 & 1 & −1 & 0 \\ 2 & 1 & 2 & 0 \\ −1 & −1 & 0 & 1 \end{matrix} | = | \begin{matrix}−1 & 2 & 0 & 1 \\ 1 & 1 & −1 & 0 \\ 2 & 1 & 2 & 0 \\ −0 & −3 & 0 & 0 \end{matrix} | .

Using a cofactor expansion about the fourth column gives

| \begin{matrix}−1 & 2 & 0 & 1 \\ 1 & 1 & −1 & 0 \\ 2 & 1 & 2 & 0 \\ −0 & −3 & 0 & 0 \end{matrix} | = (−1) | \begin{matrix}1 & 1 & −1 \\ 2 & 1 & 2 \\ 0 & 3 & 0 \end{matrix} |

= (−1)(3) | \begin{matrix}1 & −1 \\ 2 & 2 \end{matrix} | = −12.

(here, we computed the 3 × 3 determinant by expanding about the third row.

b) Using Cramer’s rule,

A^{−1}(1, 4) = \frac{C_{4,1}}{det(A)} = \frac{(−1)^{4+1} | \begin{matrix}2 & 0 & 1 \\ 1 & −1 & 0 \\ 1 & 2 & 0 \end{matrix} |}{det(A)} = \frac{−3}{−12} = \frac{1}{4}

c)

det(2A^{2}A^{T} (A^{−1})^{3}) = det(2I) det(A)^{2} det(A^{T}) det(A^{−1})^{3}
= 2^{4} · det(A)^{2} det(A) det(A)^{−3} = 16

3. A basis for the space in question is {(1, 1, 0, 0), (2, 0, 1, 0), (−1, 0, 0, 1)}. To get a orthogonal basis, we need to do the Gram-Schmidt algorithm. Start with v_{1} = (1, 1, 0, 0), v_{2} = (2, 0, 1, 0), v_{3} = (−1, 0, 0, 1),

ṽ_{1} = v_{1} = (1, 1, 0, 0)
ṽ_{2} = v_{2} − \frac{(ṽ_{1}, v_{2})}{|ṽ_{1}|^{2}} ṽ_{1} = (1, −1, 1, 0)
ṽ_{3} = v_{3} − \frac{(ṽ_{1}, v_{3})}{|ṽ_{1}|^{2}} ṽ_{1} − \frac{(ṽ_{2}, v_{3})}{|ṽ_{2}|^{2}} ṽ_{2} = (−\frac{1}{6}, \frac{1}{6}, \frac{1}{3}, 1)

Then {ṽ_{1}, ṽ_{2}, ṽ_{3}} is a set of orthogonal basis, to make them orthonormal, just multiply each ṽ_{i} by the reciprocal of its norm. So an orthonormal basis for the subspace in the problem is \{\frac{ṽ_{1}}{|ṽ_{1}|}, \frac{ṽ_{2}}{|ṽ_{2}|}, \frac{ṽ_{3}}{|ṽ_{3}|}\} = \{\frac{1}{\sqrt{2}}(1, 1, 0, 0), \frac{1}{\sqrt{3}}(1, −1, 1, 0), \frac{1}{\sqrt{42}}(−1, 1, 2, 6)\}.

4. a) A = [ \begin{matrix}1 & 0 \\ 1 & 1 \\ 1 & 4 \end{matrix} ] , x̂ = [ \begin{matrix}C \\ D \end{matrix} ] , b = [ \begin{matrix}2 \\ 3 \\ 5 \end{matrix} ], x̂ is the solution to the linear equation A^{T} Ax̂ =

A^{T} b, and the least squares line is y = C + Dx.

b) P = B(B^{T} B)^{−1}B^{T}.