Solutions
Solutions to Try Its
1. [latex]\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 1\left(-3\right)+4\left(1\right)& \hfill & \hfill 1\left(-4\right)+4\left(1\right)\\ \hfill -1\left(-3\right)+-3\left(1\right)& \hfill & \hfill -1\left(-4\right)+-3\left(1\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \\ BA=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]=\left[\begin{array}{rrr}\hfill -3\left(1\right)+-4\left(-1\right)& \hfill & \hfill -3\left(4\right)+-4\left(-3\right)\\ \hfill 1\left(1\right)+1\left(-1\right)& \hfill & \hfill 1\left(4\right)+1\left(-3\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \end{array}[/latex] 2. [latex]{A}^{-1}=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right][/latex] 3. [latex]{A}^{-1}=\left[\begin{array}{ccc}1& 1& 2\\ 2& 4& -3\\ 3& 6& -5\end{array}\right][/latex] 4. [latex]X=\left[\begin{array}{c}4\\ 38\\ 58\end{array}\right][/latex]Solutions to Odd-Numbered Exercises
1. If [latex]{A}^{-1}[/latex] is the inverse of [latex]A[/latex], then [latex]A{A}^{-1}=I[/latex], the identity matrix. Since [latex]A[/latex] is also the inverse of [latex]{A}^{-1},{A}^{-1}A=I[/latex]. You can also check by proving this for a [latex]2\times 2[/latex] matrix. 3. No, because [latex]ad[/latex] and [latex]bc[/latex] are both 0, so [latex]ad-bc=0[/latex], which requires us to divide by 0 in the formula. 5. Yes. Consider the matrix [latex]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right][/latex]. The inverse is found with the following calculation: [latex]{A}^{-1}=\frac{1}{0\left(0\right)-1\left(1\right)}\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right][/latex]. 7. [latex]AB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I[/latex] 9. [latex]AB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I[/latex] 11. [latex]AB=BA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I[/latex] 13. [latex]\frac{1}{29}\left[\begin{array}{cc}9& 2\\ -1& 3\end{array}\right][/latex] 15. [latex]\frac{1}{69}\left[\begin{array}{cc}-2& 7\\ 9& 3\end{array}\right][/latex] 17. There is no inverse 19. [latex]\frac{4}{7}\left[\begin{array}{cc}0.5& 1.5\\ 1& -0.5\end{array}\right][/latex] 21. [latex]\frac{1}{17}\left[\begin{array}{ccc}-5& 5& -3\\ 20& -3& 12\\ 1& -1& 4\end{array}\right][/latex] 23. [latex]\frac{1}{209}\left[\begin{array}{ccc}47& -57& 69\\ 10& 19& -12\\ -24& 38& -13\end{array}\right][/latex] 25. [latex]\left[\begin{array}{ccc}18& 60& -168\\ -56& -140& 448\\ 40& 80& -280\end{array}\right][/latex] 27. [latex]\left(-5,6\right)[/latex] 29. [latex]\left(2,0\right)[/latex] 31. [latex]\left(\frac{1}{3},-\frac{5}{2}\right)[/latex] 33. [latex]\left(-\frac{2}{3},-\frac{11}{6}\right)[/latex] 35. [latex]\left(7,\frac{1}{2},\frac{1}{5}\right)[/latex] 37. [latex]\left(5,0,-1\right)[/latex] 39. [latex]\frac{1}{34}\left(-35,-97,-154\right)[/latex] 41. [latex]\frac{1}{690}\left(65,-1136,-229\right)[/latex] 43. [latex]\left(-\frac{37}{30},\frac{8}{15}\right)[/latex] 45. [latex]\left(\frac{10}{123},-1,\frac{2}{5}\right)[/latex] 47. [latex]\frac{1}{2}\left[\begin{array}{rrrr}\hfill 2& \hfill 1& \hfill -1& \hfill -1\\ \hfill 0& \hfill 1& \hfill 1& \hfill -1\\ \hfill 0& \hfill -1& \hfill 1& \hfill 1\\ \hfill 0& \hfill 1& \hfill -1& \hfill 1\end{array}\right][/latex] 49. [latex]\frac{1}{39}\left[\begin{array}{rrrr}\hfill 3& \hfill 2& \hfill 1& \hfill -7\\ \hfill 18& \hfill -53& \hfill 32& \hfill 10\\ \hfill 24& \hfill -36& \hfill 21& \hfill 9\\ \hfill -9& \hfill 46& \hfill -16& \hfill -5\end{array}\right][/latex] 51. [latex]\left[\begin{array}{rrrrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill -1& \hfill -1& \hfill -1& \hfill -1& \hfill -1& \hfill 1\end{array}\right][/latex] 53. Infinite solutions. 55. 50% oranges, 25% bananas, 20% apples 57. 10 straw hats, 50 beanies, 40 cowboy hats 59. Tom ate 6, Joe ate 3, and Albert ate 3. 61. 124 oranges, 10 lemons, 8 pomegranatesLicenses & Attributions
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- Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
