Given 3x3 Matrix and 3 equations, solve the system of equations involving A

Question

Given $A\in \Bbb{R}^{3x3}$ and having $$A*e_1=e_3,\quad A*2e_3=e_1+e_3,\quad A(e_1+e_2)=e_2$$ solve: $Ax = e_1 + 2e_2 + 2e_3$

I fail to understand what the excercise expects, should I just get the $(x_1,x_2,x_3)$ in terms of those $"e"$ values? (that I assume those are vectors too because if not how that could be an equation?) My first try was just trying to get something to the right involving $A$ so I can write $$\begin{aligned} Ax &= e1+2e2+2e3 \\ Ax&=A*(\text{...})\\ A^{-1}Ax&=A^{-1}A*(\text{...}) \\ x&=(\text{...}) \end{aligned}$$ Is this the way to go? so I guess the solution after all is for any $e_1,e_2,e_3\in \Bbb{R}^3$

Answer 1

Supposing the $e_i$ denote the standard basis vectors, i. e. $e_1=(1,0,0)^T$, we get:

• If $A\cdot e_1=e_3$, we know that $A=(e_3|*|*)$.
• If $A\cdot 2e_3=2A\cdot e_3=e_1+e_3 \Leftrightarrow Ae_3=1/2(e_1+e_3)$, we know that $A=(e_3|*|1/2(e_1+e_3))$.
• If $A(e_1+e_2)=Ae_1+Ae_2=e_2 \Leftrightarrow Ae_2=e_2-Ae_1=e_2-e_3$, we know that $$A=(e_3|e_2-e_3|1/2(e_1+e_3))$$

That way, you could solve $Ax=e_1+2e_2+2e_3$ for $x$.

Answer 2

I see two ways to solve your problem. The first one, developped in Anton's answer and which you seem to want to follow, requires to determine the matrix representation of $A$, before inverting it in order to find $x = A^{-1}(e_1+2e_2+2e_3)$. The columns of the said matrix representation are actually given by the images of the basis vectors. However, it is to be noted that this method works under the assumption that the vectors $\{e_1,e_2,e_3\}$ forms indeed a basis of $\mathbb{R}^3$.

This assumption is not required with the second method, which is based on some mere substitutions in the end. We already know that $e_3 = Ae_1$ and $e_2 = A(e_1+e_2)$. Then, the second relation leads to $e_1 = 2Ae_3 - e_3 = 2A(e_3-e_1)$, hence finally $$ e_1 + 2e_2 + 2e_3 = 2A(e_3-e_1) + 2A(e_1+e_2) + 2Ae_1 = 2A(e_1+e_2+e_3) = Ax $$ and $x = 2(e_1+e_2+e_3)$.

Remark. If $A$ is not invertible, then the solution is not unique and the general solution is given by $x = 2(e_1+e_2+e_3) + y$, with $y \in \ker(A)$.