2.7.1: Exercises 2.7
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In Exercises \(\PageIndex{1}\) - \(\PageIndex{4}\), matrices \(A\) and \(B\) are given. Compute \((AB)^{-1}\) and \(B^{-1}A^{-1}\).
\(A=\left[\begin{array}{cc}{1}&{2}\\{1}&{1}\end{array}\right],\quad B=\left[\begin{array}{cc}{3}&{5}\\{2}&{5}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{-2}&{3}\\{1}&{-1.4}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right],\quad B=\left[\begin{array}{cc}{7}&{1}\\{2}&{1}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{-7/10}&{3/10}\\{29/10}&{-11/10}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{5}\\{3}&{8}\end{array}\right],\quad B=\left[\begin{array}{cc}{1}&{-1}\\{1}&{4}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{29/5}&{-18/5}\\{-11/5}&{7/5}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{4}\\{2}&{5}\end{array}\right],\quad B=\left[\begin{array}{cc}{2}&{2}\\{6}&{5}\end{array}\right]\)
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\((AB)^{-1}=\left[\begin{array}{cc}{-29/4}&{6}\\{17/2}&{-7}\end{array}\right]\)
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.
\(A=\left[\begin{array}{cc}{-3}&{5}\\{1}&{-2}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{-2}&{-5}\\{-1}&{-3}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{-3}&{5}\\{1}&{-2}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{3}&{5}\\{2}&{4}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{2}&{-5/2}\\{-1}&{3/2}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{3}&{5}\\{2}&{4}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{2}&{7}\\{1}&{3}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{-3}&{7}\\{1}&{-2}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{2}&{7}\\{1}&{3}\end{array}\right]\)
\(A=\left[\begin{array}{cc}{9}&{0}\\{7}&{9}\end{array}\right]\)
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\(A^{-1}=\left[\begin{array}{cc}{1/9}&{0}\\{-7/81}&{1/9}\end{array}\right],\quad (A^{-1})^{-1}=\left[\begin{array}{cc}{9}&{0}\\{7}&{9}\end{array}\right]\)
Find \(2\times 2\) matrices \(A\) and \(B\) that are each invertible, but \(A + B\) is not.
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Solutions will vary.
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.
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Likely some entries that should be 0 will not be exactly 0, but rather very small values
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
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Likely some entries that should be 0 will not be exactly 0, but rather very small values.