###### Exercise9

We defined an $$n\times n$$ matrix to be invertible if there is a matrix $$B$$ such that $$BA=I_n\text{.}$$ In this exercise, we will explain why $$B$$ is also invertible and that $$AB = I\text{.}$$ This means that, if $$B=A^{-1}\text{,}$$ then $$A = B^{-1}\text{.}$$

1. Given the fact that $$BA = I_n\text{,}$$ explain why the matrix $$B$$ must also be a square $$n\times n$$ matrix.

2. Suppose that $$\bvec$$ is a vector in $$\real^n\text{.}$$ Since we have $$BA = I\text{,}$$ it follows that $$B(A\bvec) = \bvec\text{.}$$ Use this to explain why the columns of $$B$$ span $$\real^n\text{.}$$ What does this say about the pivot positions of $$B\text{?}$$

3. Explain why the equation $$B\xvec = \zerovec$$ has only the trivial solution.

4. Beginning with the equation, $$BA=I\text{,}$$ multiply both sides by $$B$$ to obtain $$BAB = B\text{.}$$ We will rearrange this equation:

\begin{equation*} \begin{aligned} BAB \amp {}={} B \\ BAB - B\amp {}={} 0 \\ B(AB-I) \amp {}={} 0\text{.} \\ \end{aligned} \end{equation*}

Since the homogeneous equation $$B\xvec =\zerovec$$ has only the trivial solution, explain why $$AB-I = 0$$ and therefore, $$AB = I\text{.}$$

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