We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > Precalculus I

Solutions 33: Solving Systems with Inverses

Solutions to Try Its

1. AB=[1413][3411]=[1(3)+4(1)1(4)+4(1)1(3)+3(1)1(4)+3(1)]=[1001]BA=[3411][1413]=[3(1)+4(1)3(4)+4(3)1(1)+1(1)1(4)+1(3)]=[1001]\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} 2. A1=[35152515]{A}^{-1}=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right] 3. A1=[112243365]{A}^{-1}=\left[\begin{array}{ccc}1& 1& 2\\ 2& 4& -3\\ 3& 6& -5\end{array}\right] 4. X=[43858]X=\left[\begin{array}{c}4\\ 38\\ 58\end{array}\right]

Solutions to Odd-Numbered Exercises

1. If A1{A}^{-1} is the inverse of AA, then AA1=IA{A}^{-1}=I, the identity matrix. Since AA is also the inverse of A1,A1A=I{A}^{-1},{A}^{-1}A=I. You can also check by proving this for a 2×22\times 2 matrix. 3. No, because adad and bcbc are both 0, so adbc=0ad-bc=0, which requires us to divide by 0 in the formula. 5. Yes. Consider the matrix [0110]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]. The inverse is found with the following calculation: A1=10(0)1(1)[0110]=[0110]{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]. 7. AB=BA=[1001]=IAB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I 9. AB=BA=[1001]=IAB=BA=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I 11. AB=BA=[100010001]=IAB=BA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I 13. 129[9213]\frac{1}{29}\left[\begin{array}{cc}9& 2\\ -1& 3\end{array}\right] 15. 169[2793]\frac{1}{69}\left[\begin{array}{cc}-2& 7\\ 9& 3\end{array}\right] 17. There is no inverse 19. 47[0.51.510.5]\frac{4}{7}\left[\begin{array}{cc}0.5& 1.5\\ 1& -0.5\end{array}\right] 21. 117[55320312114]\frac{1}{17}\left[\begin{array}{ccc}-5& 5& -3\\ 20& -3& 12\\ 1& -1& 4\end{array}\right] 23. 1209[475769101912243813]\frac{1}{209}\left[\begin{array}{ccc}47& -57& 69\\ 10& 19& -12\\ -24& 38& -13\end{array}\right] 25. [1860168561404484080280]\left[\begin{array}{ccc}18& 60& -168\\ -56& -140& 448\\ 40& 80& -280\end{array}\right] 27. (5,6)\left(-5,6\right) 29. (2,0)\left(2,0\right) 31. (13,52)\left(\frac{1}{3},-\frac{5}{2}\right) 33. (23,116)\left(-\frac{2}{3},-\frac{11}{6}\right) 35. (7,12,15)\left(7,\frac{1}{2},\frac{1}{5}\right) 37. (5,0,1)\left(5,0,-1\right) 39. 134(35,97,154)\frac{1}{34}\left(-35,-97,-154\right) 41. 1690(65,1136,229)\frac{1}{690}\left(65,-1136,-229\right) 43. (3730,815)\left(-\frac{37}{30},\frac{8}{15}\right) 45. (10123,1,25)\left(\frac{10}{123},-1,\frac{2}{5}\right) 47. 12[2111011101110111]\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] 49. 139[3217185332102436219946165]\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] 51. [100000010000001000000100000010111111]\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] 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 pomegranates

Licenses & Attributions