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Suppose that $A$ has no inverse. Prove that there exists a vector $b$ such that $Ax = b$ has no solution

My try

Proving by contradiction , Assume that for all vector $b$, $Ax = b$ have at least one solution. This implies

$$A^{-1}Ax = A^{-1}b$$ which then gives $$x=A^{-1}b$$ but this means that $A^{-1}$ exists and hence contradicting our assumption that $A$ has no inverse, hence proving our original statement. Is my prove correct.Could anyone explain Thanks. Also lets say if i would to use a direct proof instead. How would i go about doing it.

ys wong
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3 Answers3

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Use the contrapositive.

If $Ax=b$ has at least one solution for each $b$, then you can solve $Ax=e_i$ for each $e_i$, where $e_i$ is a vector in the canonical basis. But then the solutions $x_i$ can be put as columns in a matrix $X$ and get $AX=I$. We have then found an inverse for $A$. (At least, a right inverse.)

lhf
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If the determinant is zero then the rows and columns of A are linearly dependent! look at this: What does it mean to have a determinant equal to zero?

This means that the sets of equations you get from $Ax=b$ are linearly dependent! Then the number of equations $Ax$ might be less than the number unknowns $b$! In which case a solution does not exist!

Community
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Ben
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Elementary matrices can be used for a proof. If $A$ is not invertible, then there are elementary matrices $E_r,...,E_1$ such that $$E_r...E_1A=A_{red}$$ where the reduced matrix $A_{red}$'s bottom row is zero row. Let $\vec v=[0\;0 ...0\;0\;1]^{T}$. Then $A_{red}\vec x=\vec v$ has no solution. Hence, $$E_r...E_1A\vec x=\vec v$$ $$A\vec x=E_1^{-1}...E_r^{-1}\vec v=\vec {w}$$ has no solution.

Bob Dobbs
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