\( \newcommand{\blah}{blah-blah-blah} \newcommand{\eqb}[1]{\begin{eqnarray*}#1\end{eqnarray*}} \newcommand{\eqbn}[1]{\begin{eqnarray}#1\end{eqnarray}} \newcommand{\bb}[1]{\mathbf{#1}} \newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}} \newcommand{\nchoose}[2]{\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} \newcommand{\defn}{\stackrel{\vartriangle}{=}} \newcommand{\rvectwo}[2]{\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} \newcommand{\rvecthree}[3]{\left(\begin{array}{r} #1 \\ #2\\ #3\end{array}\right)} \newcommand{\rvecdots}[3]{\left(\begin{array}{r} #1 \\ #2\\ \vdots\\ #3\end{array}\right)} \newcommand{\vectwo}[2]{\left[\begin{array}{r} #1\\#2\end{array}\right]} \newcommand{\vecthree}[3]{\left[\begin{array}{r} #1 \\ #2\\ #3\end{array}\right]} \newcommand{\vecdots}[3]{\left[\begin{array}{r} #1 \\ #2\\ \vdots\\ #3\end{array}\right]} \newcommand{\eql}{\;\; = \;\;} \definecolor{dkblue}{RGB}{0,0,120} \definecolor{dkred}{RGB}{120,0,0} \definecolor{dkgreen}{RGB}{0,120,0} \)

Assignment 2


 

Part I: Pen-and-paper

  1. Suppose that the inverse of a square matrix is defined using only left multiplication: define \({\bf A}^{-1}\) as the matrix with the property that \({\bf A}^{-1} {\bf A} = {\bf I}\). In this exercise, you will prove that this implies that \({\bf A} {\bf A}^{-1} = {\bf I}\) using two steps:
    • First show that the identity matrix \({\bf I}\) is unique. That is, if there's any other matrix \({\bf J}\) such that \({\bf A J} = {\bf A} = {\bf J A}\), then \({\bf I} = {\bf J}\).
    • Use the above fact and the associativity of matrix multiplication (which is a hint too) to establish \({\bf A} {\bf A}^{-1} = {\bf I}\).


Part II: Programming
The next set of functionality to build into your library is to solve equations.



Submission