Matrix equation Ax=b is A nonsingular when $b \in \mathbb{R}^n$?

Question

Problem is that i dont know if this is good solution to this problem or if i am missing something ? I would like to have some feedback.

Question:

We have matrix equation $Ax=b$ where $A$ is $n \times n$ matrix and equation $Ax=b$ has solution when $b \in \mathbb{R}^n$. Is matrix $A$ Nonsingular ? Provide explanation with answer.

Attempt to solve:

Matrix $A$ is nonsingular if Determinant is not zero.

$$Det(A)=|A| \neq 0$$

Meaning if $Det(A)\neq 0$ we have possibility to calculate inverse matrix for matrix $A$. There most be such $n \times n$-matrix $B$ that,

$$ AB=BA=I_{n} $$ $$ B=A^{-1} $$ Where $I_n$ is identity matrix. Meaning a singularmatrix that has value 1 in diagonal line and value of 0 elsewhere. Identity matrix is also ortogonal. $$\\$$ Equation Ax=b has solution if, $$ Ax=b $$ $$ x=A^{-1}b $$ In order to solve $Ax=b$ there has to be inverse matrix for $A$. Matrix $A$ has to be nonsingular if equation $Ax=b$ has every possible solution when $b \in \mathbb{R}^n$ $$\\$$ Any comment providing feedback would be much appreciated.

Thanks,

Tuki

Answer 1

To show that if $Ax=b$ has solution for all $b$, then $A$ must be non-singular:

The equations $$Ax=e_i$$ has solution $x_i.$ Then $$A(x_1, x_2, \cdots, x_n)=(e_1, e_2, \cdots, e_n)=I$$ and hence $A$ has inverse and is non-singular.

Answer 2

Since $Ax=b$ has solution for all $b \in \mathbb{R}^n$, the linear operator $L_A \colon \mathbb{R}^n \to \mathbb{R}^n$ given by $L_A(x):=Ax$ is surjective.

Therefore by the Rank-Nullity theorem it is must be also injective, so necessarily $\det A \neq 0$.