Skip to main content
Mathematics LibreTexts

2.7.1: Exercises 2.7

  • Page ID
    69518
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    In Exercises \(\PageIndex{1}\) - \(\PageIndex{4}\), matrices \(A\) and \(B\) are given. Compute \((AB)^{-1}\) and \(B^{-1}A^{-1}\).

    Exercise \(\PageIndex{1}\)

    \(A=\left[\begin{array}{cc}{1}&{2}\\{1}&{1}\end{array}\right],\quad B=\left[\begin{array}{cc}{3}&{5}\\{2}&{5}\end{array}\right]\)

    Answer

    \((AB)^{-1}=\left[\begin{array}{cc}{-2}&{3}\\{1}&{-1.4}\end{array}\right]\)

    Exercise \(\PageIndex{2}\)

    \(A=\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right],\quad B=\left[\begin{array}{cc}{7}&{1}\\{2}&{1}\end{array}\right]\)

    Answer

    \((AB)^{-1}=\left[\begin{array}{cc}{-7/10}&{3/10}\\{29/10}&{-11/10}\end{array}\right]\)

    Exercise \(\PageIndex{3}\)

    \(A=\left[\begin{array}{cc}{2}&{5}\\{3}&{8}\end{array}\right],\quad B=\left[\begin{array}{cc}{1}&{-1}\\{1}&{4}\end{array}\right]\)

    Answer

    \((AB)^{-1}=\left[\begin{array}{cc}{29/5}&{-18/5}\\{-11/5}&{7/5}\end{array}\right]\)

    Exercise \(\PageIndex{4}\)

    \(A=\left[\begin{array}{cc}{2}&{4}\\{2}&{5}\end{array}\right],\quad B=\left[\begin{array}{cc}{2}&{2}\\{6}&{5}\end{array}\right]\)

    Answer

    \((AB)^{-1}=\left[\begin{array}{cc}{-29/4}&{6}\\{17/2}&{-7}\end{array}\right]\)

    In Exercises \(\PageIndex{5}\) - \(\PageIndex{8}\), a \(2\times 2\) matrix \(A\) is given. Compute \(A^{-1}\) and \((A^{-1})^{-1}\) using Theorem 2.6.3.

    Exercise \(\PageIndex{5}\)

    \(A=\left[\begin{array}{cc}{-3}&{5}\\{1}&{-2}\end{array}\right]\)

    Answer

    \(A^{-1}=\left[\begin{array}{cc}{-2}&{-5}\\{-1}&{-3}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{-3}&{5}\\{1}&{-2}\end{array}\right]\)

    Exercise \(\PageIndex{6}\)

    \(A=\left[\begin{array}{cc}{3}&{5}\\{2}&{4}\end{array}\right]\)

    Answer

    \(A^{-1}=\left[\begin{array}{cc}{2}&{-5/2}\\{-1}&{3/2}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{3}&{5}\\{2}&{4}\end{array}\right]\)

    Exercise \(\PageIndex{7}\)

    \(A=\left[\begin{array}{cc}{2}&{7}\\{1}&{3}\end{array}\right]\)

    Answer

    \(A^{-1}=\left[\begin{array}{cc}{-3}&{7}\\{1}&{-2}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{2}&{7}\\{1}&{3}\end{array}\right]\)

    Exercise \(\PageIndex{8}\)

    \(A=\left[\begin{array}{cc}{9}&{0}\\{7}&{9}\end{array}\right]\)

    Answer

    \(A^{-1}=\left[\begin{array}{cc}{1/9}&{0}\\{-7/81}&{1/9}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{9}&{0}\\{7}&{9}\end{array}\right]\)

    Exercise \(\PageIndex{9}\)

    Find \(2\times 2\) matrices \(A\) and \(B\) that are each invertible, but \(A + B\) is not.

    Answer

    Solutions will vary.

    Exercise \(\PageIndex{10}\)

    Create a random \(6\times 6\) matrix \(A\), then have a calculator or computer compute \(AA^{−1}\). Was the identity matrix returned exactly? Comment on your results.

    Answer

    Likely some entries that should be 0 will not be exactly 0, but rather very small values

    Exercise \(\PageIndex{11}\)

    Use a calculator or computer to compute \(AA^{-1}\), where

    \[A=\left[\begin{array}{cccc}{1}&{2}&{3}&{4}\\{1}&{4}&{9}&{16}\\{1}&{8}&{27}&{64}\\{1}&{16}&{81}&{256}\end{array}\right]. \nonumber \]

    Was the identity matrix returned exactly. Comment on your results.

    Answer

    Likely some entries that should be 0 will not be exactly 0, but rather very small values.


    2.7.1: Exercises 2.7 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?