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
Best Answer
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.