MHB Range and Image of a Transformation

[bwpbruce]

$\textbf{Problem}$

Let $\textbf{b} = \begin{bmatrix}\begin{array}{r} 8 \\ 7 \\ 5 \\ -3 \end{array}\end{bmatrix}$ and let $A = \begin{bmatrix} 2 & 3 & 5 & - 5 \\ -7 & 7 & 0 & 0 \\ -3 & 4 & 1 & 3 \\ -9 & 3 & -6 & -4 \end{bmatrix}$

Is $\textbf{b}$ in the range of the transformation $\textbf{x} \mapsto A\textbf{x}$. If so, find an $\textbf{x}$ whose image under the transformation is $\textbf{b}$.$\textbf{My Solution}$
By inspection, notice that the sum of columns $\textbf{a}_2$ and $\textbf{a}_3$ is $\textbf{b}$.

In other words: \begin{align*}\begin{bmatrix} 3 \\ 7 \\ 4 \\ 3 \end{bmatrix} + \begin{bmatrix} 5 \\ 0 \\ 1 \\ -6 \end{bmatrix} &= \begin{bmatrix} 3 + 5 \\ 7 + 0 \\ 4 + 1 \\ 3 - 6 \end{bmatrix} = \begin{bmatrix} 8 \\ 7 \\ 5 \\ -3 \end{bmatrix} \end{align*}
So $\textbf{b}$ is in the range of $T$.
Also by inspection, it is obvious that for $A\textbf{(x)} = \textbf{b}$, $\textbf{x} = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}$.

Is my approach to this acceptable?

 

[Evgeny.Makarov]

Is the element of $A$ in row 3, column 2 equal to $-4$ or $4$? If it is $-4$, then $\mathbf{a}_2+\mathbf{a}_3\ne\mathbf{b}$.

If that element is 4, then your solution is correct, but what would you do if you couldn't guess an answer?

 

[bwpbruce]

Is the element of $A$ in row 3, column 2 equal to $-4$ or $4$? If it is $-4$, then $\mathbf{a}_2+\mathbf{a}_3\ne\mathbf{b}$.

If that element is 4, then your solution is correct, but what would you do if you couldn't guess an answer?

You're good at finding typos. What do you mean by if you $\textbf{couldn't}$ guess an answer? I think I know this "other way" you're referring to. But I'm just saying... Show me a situation where $\textbf{b}$ is in the range but one is forced to use the alternative method rather than the method I posted above.

 

[Evgeny.Makarov]

Let $A$ be the same and $\mathbf{b}=\begin{bmatrix}\dfrac{151}{6}\\ -21\\ -\dfrac{37}{2}\\ -\dfrac{50}{3}\end{bmatrix}$. Is $\mathbf{b}$ in the range of $\mathbf{x}\mapsto A\mathbf{x}$?

 

[bwpbruce]

That's clever. The alternative method is to create a matrix of the form $\begin{bmatrix}A & \textbf{b} \end{bmatrix}$ then simplify to reduced echelon form. If the system is consistent then $\textbf{b}$ is in the range of $\textbf{x} \mapsto A\textbf{x}$.

 

[Evgeny.Makarov]

Yes.

 

[bwpbruce]

Well, for the given problem, my method was more efficient.