18.06 Exam 1 #1 Solutions

1 a)

v⃗ · x⃗ = 0 ⇒ x_{1} + 2x_{2} + x_{3} = 0
w⃗ · x⃗ = 0 ⇒ 2x_{1} + 4x_{2} + 3x_{3} = 0

So the set to be found is the nullspace of the matrix A = [ \begin{matrix}1 & 2 & 1 \\ 2 & 4 & 3 \end{matrix} ]. The row echelon form

of A is [ \begin{matrix}1 & 2 & 1 \\ 0 & 0 & 1 \end{matrix} ]. The second variable, x_{2}, is free and the vector (−2, 1, 0) is a basis of the

nullspace.

b) Since the set in a) is the nullspace of the matrix A, it is a vector space. Generally to prove a set satisfying some property, say P, is a vector space, one needs to show:

(1) If x⃗ satisfies property P, then cx⃗ also satisfies property P, for any c ∈ ℝ.

(2) If x⃗, y⃗ satisfy property P, then x⃗ + y⃗ also satisfies property P. 2 a)

A = [ \begin{matrix}−2 & 0 & 3 \\ −4 & 3 & −2 \\ 8 & 9 & 11 \end{matrix} ] = [ \begin{matrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ −4 & 0 & 1 \end{matrix} ] · [ \begin{matrix}−2 & 0 & 3 \\ 0 & 3 & −8 \\ 0 & 9 & 23 \end{matrix} ]

= [ \begin{matrix}1 & 0 & 0 \\ −2 & 1 & 0 \\ 4 & 0 & 1 \end{matrix} ] · [ \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 3 & 1 \end{matrix} ] · [ \begin{matrix}−2 & 0 & 3 \\ 0 & 3 & −8 \\ 0 & 0 & 47 \end{matrix} ]

So

L = [ \begin{matrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ −4 & 0 & 1 \end{matrix} ] · [ \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 3 & 1 \end{matrix} ] = [ \begin{matrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ −4 & 3 & 1 \end{matrix} ]

U = [ \begin{matrix}−2 & 0 & 3 \\ 0 & 3 & −8 \\ 0 & 0 & 47 \end{matrix} ]

b) To solve Ax⃗ = LUx⃗ = b⃗, it is equivalent to solve the two equations Ly⃗ = b⃗ and Ux⃗ = y⃗.

Ly⃗ = b⃗ ⇒ \{ \begin{matrix}y_{1} & = & 3 \\ 2y_{1} + y_{2} & = & −1 \\ −4y_{1} + 3y_{2} + y_{3} & = & 13 \end{matrix} ⇒ y⃗ = [ \begin{matrix}3 \\ −7 \\ 46 \end{matrix} ] .

Ux⃗ = y⃗ ⇒ y⃗ = [ \frac{−3}{94} ; \frac{13}{47} ; \frac{46}{47} ] .

3 a) Denote B = [ \begin{matrix}1 & 2 & 4 \\ 3 & 1 & 7 \\ 5 & 2 & 6 \end{matrix} ] , P = [ \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ], Then

BA = P ⇒ P^{−1}BA = I ⇒ P^{−1}B = A^{−1}

Because P is a permutation matrix, P^{−1} = P^{T}. So

A^{−1} = [ \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} ] · [ \begin{matrix}1 & 2 & 4 \\ 3 & 1 & 7 \\ 5 & 2 & 6 \end{matrix} ] = [ \begin{matrix}3 & 1 & 7 \\ 5 & 2 & 6 \\ 1 & 2 & 4 \end{matrix} ] .

b) i B, D have full column rank, so the nullspace of each is the zero vector. Now

BDx⃗ = 0 ⇒ Dx⃗ ∈ N(B) = {0} ⇒ Dx⃗ = 0 ⇒ x⃗ = 0.

Hence N(BD)=0. ii This time only B has full column rank, that is, N(B) = {0}.

BDx⃗ = 0 ⇒ Dx⃗ ∈ N(B) = {0} ⇒ Dx⃗ = 0 ⇒ x⃗ ∈ N(D).

So N(BD) ⊆ N(D). On the other hand,

Dx⃗ = 0 ⇒ BDx⃗ = B0 = 0 ⇒ x ∈ N(BD) ⇒ N(D) ⊆ N(BD).

So N(D) = N(BD), which is all we can say about N(BD) without further assumptions on D.

iii r < n, implies B is not of full column rank and the nullspace of B contains an infinite number of vectors. r < m implies the row echelon form of B has zero rows, so the equation Bx⃗ = b⃗ has no solutions for some b⃗. Furthermore, if there is a solution to Bx⃗ = b⃗, say x⃗_{p}, then there are infinitely many solutions since x⃗_{p} + x⃗_{n} is a solution for any x⃗_{n} in N(B). The answer to the question is 0 or infinitely many.

4 a) Apply row operations on A and get the following matrix

R = [ \begin{matrix}1 & 2 & 3 & 4 \\ 0 & −1 & c − 3 & −3 \\ 0 & 0 & 2(c − 3) & −8 \\ 0 & 0 & 0 & d − 8 \end{matrix} ] .

– No values of c, d will make the rank of A equal to 2.

– if c ≠ 3, d ≠ 8, R is the row echelon form of A and A has rank 4.

– Any other combination of c, d will give rank 3, that is, the rank is 3 if c = 3 or d = 8.

b) substituting c = 3, d = 8 in the matrix R, one finds that the third column gives a free variable, and null space of A is spanned by (−3, 0, 1, 0). Use the augmented matrix [A | b⃗] (NOT

[R | b⃗]) to find a particular solution of the equation Ax⃗ = b⃗, which is (−1/2, 1/4, 0, 1/4). So the complete solution of the equation is (−1/2, 1/4, 0, /1/4) + x_{3}(−3, 0, 1, 0).