Geometric/vector explanation of $\det(A)=0\iff$ unique solution doesn't exist to system of linear equation Currently, I am self-studying Multivariable Calculus. I have prior knowledge about Vectors, Matrices and System of Linear Equations. However, the linkages between the three are not explicitly covered by my prior education and curriculum. I reckon it is beneficial for me to build a correct understanding about these linkages.
I am not most familiar with the language of Mathematics. I apologize for any inaccuracies or intracies throughout my question in advance.

TL;DR: Explain the following geometrically / from "vector view":$$\begin{align}&\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are linearly dependent}\\\impliedby&\overrightarrow{d}\text{ cannot be expressed in any linear combination of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\end{align}$$

Suppose there is a system of linear equations in the forms $$x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c}=\overrightarrow{d}$$
and $$\mathbf{A}\overrightarrow{x}=\overrightarrow{d}$$
From my understanding, solving this system of linear equations is conceptually equivalent to:

*

*in a "vector view", expressing $\overrightarrow{d}$ in terms of linear combination(s) of $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$;

*in a "matrix view", solving $\overrightarrow{x}=\mathbf{A}\overrightarrow{d}$.

My goal is to understand "$\det(\mathbf{A}^{-1})=0\iff\text{unique solution doesn't exist}$" from the "vector view".
I believe there is a logical explanation to my question: $$\begin{align}&\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are linearly dependent}\\\iff&\det\mathbf{A}=0\\\iff&\text{unique solution doesn't exist}\\\iff&\text{no solution or infinitely many solutions}\\\iff&\overrightarrow{d}\text{ cannot be expressed in any linear combination of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\\&\text{ or there exist infinitely many linear combinations of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ expressing }\overrightarrow{d}\end{align}$$
To put it simply, the logical explanation is $$\begin{align}&\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are linearly dependent}\\\iff&\overrightarrow{d}\text{ cannot be expressed in any linear combination of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\\&\text{ or there exist infinitely many linear combinations of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ expressing }\overrightarrow{d}\end{align}$$
From the definition of basis vectors, I know that $$\begin{align}&\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are linearly independent}\\\iff&\overrightarrow{d}\text{ can be expressed in }\mathbf{unique }\text{ linear combination of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\end{align}$$
and I understand that the negation of the above statement would give the previous statement. **My understanding is stuck in the fact that **$$\begin{align}&\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are linearly dependent}\\\impliedby&\overrightarrow{d}\text{ cannot be expressed in any linear combination of }\overrightarrow{a}\text{, }\overrightarrow{b}\text{ and }\overrightarrow{c}\end{align}$$
$\det{A}=0\impliedby\text{no solution}$ is the reason why the above statement is true. I lack the intuition that it is true. Could anyone fill me in with the "vector view" of such statement?
Thank you.
 A: I understand you want to try: Let $a,bc\in\mathbb R^3$ nonzero vectors.
If for some $x\in\mathbb R^3$ non exists $\alpha,\beta,\gamma\in\mathbb R$ such that $x=\alpha\cdot a+\beta\cdot b+\gamma\cdot c$. Then $a,b$ and $c$ are linearly dependent vectors.
or equivalently, we have:
If $a,b$ and $c$ are linearly independent vector. Then for all $x\in\mathbb R^3$ exists $\alpha,\beta,\gamma\in\mathbb R$ such that $x=\alpha\cdot a+\beta\cdot b+\gamma\cdot c$.
Here is a solution with vector spaces
Since $a,b$ and $c$ are linearly independent vector. Let $Span\{a,b,c\}=$set of all linear combinations of the vectors a, b, and c. Then $\dim(Span\{a,b,c\})=3$ and $\dim\mathbb R^3=3$. Hence $\mathbb R^3=Span\{a,b,c\}$. And therefore let be any vector $x\in\mathbb R^3=Span\{a,b,c\}$, i.e. $x$ is linear combinations of the vector $a,b$ and $c$.
I am making use of the following result: Let $F$ be a vector subspace of an n-dimensional vector space $E$. If $\dim F=n$ then $F=E$.
Here you have a solution with matrices
Since $a,b$ and $c$ are linearly independent vector. Let $M=\begin{pmatrix}
a_1&b_1&c_1\\
a_2&b_2&c_2\\
a_3&b_3&c_3
\end{pmatrix}$ with $\det(M)\neq 0$. We want to solve the following system
$$
\begin{pmatrix}
a_1&b_1&c_1\\
a_2&b_2&c_2\\
a_3&b_3&c_3
\end{pmatrix}\cdot\begin{pmatrix}
\alpha\\
\beta\\
\gamma
\end{pmatrix}=\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}\Rightarrow \begin{pmatrix}
\alpha\\
\beta\\
\gamma
\end{pmatrix}=\begin{pmatrix}
a_1&b_1&c_1\\
a_2&b_2&c_2\\
a_3&b_3&c_3
\end{pmatrix}^{-1}\cdot \begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$
It is so that it has a unique solution. That is for all $x\in\mathbb R^3$ exists $\alpha,\beta,\gamma\in\mathbb R$ such that $x=\alpha\cdot a+\beta\cdot b+\gamma\cdot c$.
I hope I have understood your question well.
