Help with $(iii)$ of Math Problem on MHB

[Guest2]

Greetings, the good people of MHB! I know that the answer to $(i)$ is $(D)$; and since the third row contains $0=1$, the system has no solutions, and their intersection is the empty set, so the answer to $(ii)$ is $(D)$. I'd appreciate any help with the rest of the question.

For $(iii)$ I get $\mathbf{x} = x_{3}(-2,1,1,0)^{T}$, so the dimension is $1$, a line through the origin, so I think, though not too sure, the answer is $(F)$.

 

[Jameson]

Hi Guest,

Let me preface this with my linear algebra is not great, but until a better helper can assist you I'll try. :)

I got the same thing for the null space and agree that this will plot a line through the origin. For that reason I think it comes down to is this a line in $\mathbb{R}^3$ or $\mathbb{R}^4$? So if that's the case it would be either B or E right?

For iv, since this is a linear map I believe we can break up $T(a+b)=T(a)+T(b)$ and consider each of these separately.

 

[Deveno]

Hi Guest,

Let me preface this with my linear algebra is not great, but until a better helper can assist you I'll try. :)

I got the same thing for the null space and agree that this will plot a line through the origin. For that reason I think it comes down to is this a line in $\mathbb{R}^3$ or $\mathbb{R}^4$? So if that's the case it would be either B or E right?

For iv, since this is a linear map I believe we can break up $T(a+b)=T(a)+T(b)$ and consider each of these separately.

The null space is always a subspace of the "domain" of the matrix (what size vectors we can multiply $A$ on the left by). This is equal to the number of columns $A$ has, as these must match the number of "rows" (entries) our column vectors have. This should settle (iii).

Your suggestion for (iv) is indeed the way to go, what good is studying linearity, if we cannot leverage it to do the heavy lifting?

 

[Guest2]

Many thanks, guys.

So for my own reference, so that I can come back to it come exam time.

(iii) is (E) a line in $
\mathbb{R}^4$

(iv) $T$ is a map from $\mathbb{R}^4$ to $\mathbb{R}^3$, so we compute $\mathbf{A} = \begin{pmatrix} 2&-1&5&8 \\ 2&4&0&5 \\ 1&3&-1&4
\end{pmatrix} (3,-2,-1,0)^{T} = (-2,-2,-1) = -2\mathbf{e}_1-2\mathbf{e}_2-\mathbf{e}_3$

So the answer is $(F)$, none of these.

(v) We row reduce $\left(\begin{array}{cccc|c} 2&-1&5&8&3 \\ 2&4&0&5&-2 \\ 1&3&-1&4&-2
\end{array}\right) \to \left(\begin{array}{cccc|c} 1&0&2&0&1 \\ 0&1&-1&0&-1 \\ 0&0&0&1&0
\end{array}\right)$

So the solutions are $(x_1, x_2, x_4) = (1-2x_{3}, x_3-1, 0)$