Find image of matrix $A$

Question

Let $A=\begin{pmatrix}7&5&1\\-8&-6&0\\-3&-2&-1\end{pmatrix}$ find Im(A).

My attempt:

We know that $Im(A)=\{B=\begin{pmatrix}a\\b\\c\end{pmatrix}\in\mathbb{R}^3|Ax=B, x\in \mathbb{R^3}\}.$

So then I try solving it with Gauss:

$\begin{array}{l}\begin{bmatrix}7&5&1&\vdots&a\\-8&-6&0&\vdots&b\\-3&-2&-1&\vdots&c\end{bmatrix}\xrightarrow[{L_3=-L_3+\frac{-3}7L_1}]{L_2=\frac{-1}2L2+\frac{-4}7L_1}\\\begin{bmatrix}7&5&1&\vdots&a\\0&\frac17&-\frac47&\vdots&-\frac b2-\frac47a\\0&-\frac17&\frac47&\vdots&-c-\frac37a\end{bmatrix}\xrightarrow[{L_3=7L_3+L_2}]{L_2=7L_2}\\\begin{bmatrix}7&5&1&\vdots&a\\0&1&-4&\vdots&-\frac72b-4a\\0&0&0&\vdots&-7c-7a-\frac72b\end{bmatrix}\end{array}$

then from the last equation we get: $\boxed{a=-c-\frac{b}{2}}$

so now.... what will my vector $B$ will be? Because if I write it in this form I will get a 0 at the bottom and that's not good

Answer 1

There is a problem with your answer: the symbol $b$ has two distinct meanings.

Anyway, assuming that your computations are correct (I didn't check it), the conclusion is that $$\operatorname{Im}(A)=\left\{(a,b,c)\in\mathbb{R}^3\,\middle|\,a=-c-\frac b2\right\}=\left\{\left(-c-\frac b2,b,c\right)\,\middle|\,b,c\in\mathbb R\right\}.$$

Answer 2

The image is determined when you have a set of vectors which span it.

The columns of the matrix $A$ are a spanning set for the image, as these are the images of the unit vectors $(1,0,0), (0,1,0), (0,0,1)$.

If you apply Gaussian elimination to these columns (rather than to the rows, as you have done), you will find the basis: $(1,0,-1)^T, (0,2,-1)^T$. These define the plane $x + \frac{y}2 +z = 0$ supplied by Juan Carlos Santos in his answer above.