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Please explain this in the simplest terms possible! Thank you in advance!

If $A$ is a $3\times3$ matrix and $$A\mathbf x = \begin{bmatrix} 1\\-2\\7\end{bmatrix}$$ has exactly one solution, explain why $A$ must be invertible.

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2 Answers 2

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We are going to show that $A$ has maximum rank, in particular trivial kernel, it will follow that $A$ is also invertible. Suppose by contradiction that there exists an $y \neq 0$ such that $Ay=0$, it follows: $$ A(x+y)=Ax+Ay= \begin{bmatrix} 1\\-2\\7\end{bmatrix}.$$ Hence $x+y$ is also a solution, absurd. So we must have $x+y=x \implies y=0$ and $\text{Ker}(A)=\{ \mathbf{0}\}$.

A further idea: now we know that $A$ has full rank, by Rouchè Capelli it follows that for every $b \neq 0$, the linear equation $Ax=b$ has exactly one solution. We can directly construct the inverse: let $(\mathbf{e}_i)_{i=1,2,3}$ be the canonical base of $\mathbb{R}^3$ (the column vectors of the identity matrix), since all $\mathbf{e}_i$ are non-zero, the system $Ax=\mathbf{e}_i$ has exactly one solution for $i=1,2,3$. With this information you should be able to construct $A^{-1}$ (just think about matrix multiplication).

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If $A$ is not invertible, then its kernel is not trivial.

Let $u\ne0$ be in the kernel of $A$. Let $v$ be any solution for $Ax=b$.

Then $u+v$ is another solution of $Ax=b$.