{"id":1810,"date":"2015-11-12T18:30:44","date_gmt":"2015-11-12T18:30:44","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1810"},"modified":"2015-11-12T18:30:44","modified_gmt":"2015-11-12T18:30:44","slug":"solutions-21","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/chapter\/solutions-21\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions to Try Its<\/h2>\n1.\u00a0[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]\n\n2. [latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}& \\frac{1}{5}\\\\ -\\frac{2}{5}& \\frac{1}{5}\\end{array}\\right][\/latex]\n\n3.\u00a0[latex]{A}^{-1}=\\left[\\begin{array}{ccc}1& 1& 2\\\\ 2& 4& -3\\\\ 3& 6& -5\\end{array}\\right][\/latex]\n\n4.\u00a0[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]\n

Solutions to Odd-Numbered Exercises<\/h2>\n1.\u00a0If [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.\n\n3.\u00a0No, 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.\n\n5.\u00a0Yes. 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].\n\n7.\u00a0[latex]AB=BA=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]=I[\/latex]\n\n9.\u00a0[latex]AB=BA=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]=I[\/latex]\n\n11.\u00a0[latex]AB=BA=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]=I[\/latex]\n\n13.\u00a0[latex]\\frac{1}{29}\\left[\\begin{array}{cc}9& 2\\\\ -1& 3\\end{array}\\right][\/latex]\n\n15.\u00a0[latex]\\frac{1}{69}\\left[\\begin{array}{cc}-2& 7\\\\ 9& 3\\end{array}\\right][\/latex]\n\n17.\u00a0There is no inverse\n\n19.\u00a0[latex]\\frac{4}{7}\\left[\\begin{array}{cc}0.5& 1.5\\\\ 1& -0.5\\end{array}\\right][\/latex]\n\n21.\u00a0[latex]\\frac{1}{17}\\left[\\begin{array}{ccc}-5& 5& -3\\\\ 20& -3& 12\\\\ 1& -1& 4\\end{array}\\right][\/latex]\n\n23.\u00a0[latex]\\frac{1}{209}\\left[\\begin{array}{ccc}47& -57& 69\\\\ 10& 19& -12\\\\ -24& 38& -13\\end{array}\\right][\/latex]\n\n25.\u00a0[latex]\\left[\\begin{array}{ccc}18& 60& -168\\\\ -56& -140& 448\\\\ 40& 80& -280\\end{array}\\right][\/latex]\n\n27.\u00a0[latex]\\left(-5,6\\right)[\/latex]\n\n29.\u00a0[latex]\\left(2,0\\right)[\/latex]\n\n31.\u00a0[latex]\\left(\\frac{1}{3},-\\frac{5}{2}\\right)[\/latex]\n\n33.\u00a0[latex]\\left(-\\frac{2}{3},-\\frac{11}{6}\\right)[\/latex]\n\n35.\u00a0[latex]\\left(7,\\frac{1}{2},\\frac{1}{5}\\right)[\/latex]\n\n37.\u00a0[latex]\\left(5,0,-1\\right)[\/latex]\n\n39.\u00a0[latex]\\frac{1}{34}\\left(-35,-97,-154\\right)[\/latex]\n\n41.\u00a0[latex]\\frac{1}{690}\\left(65,-1136,-229\\right)[\/latex]\n\n43.\u00a0[latex]\\left(-\\frac{37}{30},\\frac{8}{15}\\right)[\/latex]\n\n45.\u00a0[latex]\\left(\\frac{10}{123},-1,\\frac{2}{5}\\right)[\/latex]\n\n47.\u00a0[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]\n\n49.\u00a0[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]\n\n51.\u00a0[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]\n\n53.\u00a0Infinite solutions.\n\n55.\u00a050% oranges, 25% bananas, 20% apples\n\n57.\u00a010 straw hats, 50 beanies, 40 cowboy hats\n\n59.\u00a0Tom ate 6, Joe ate 3, and Albert ate 3.\n\n61.\u00a0124 oranges, 10 lemons, 8 pomegranates","rendered":"

Solutions to Try Its<\/h2>\n

1.\u00a0[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]<\/p>\n

2. [latex]{A}^{-1}=\\left[\\begin{array}{cc}\\frac{3}{5}& \\frac{1}{5}\\\\ -\\frac{2}{5}& \\frac{1}{5}\\end{array}\\right][\/latex]<\/p>\n

3.\u00a0[latex]{A}^{-1}=\\left[\\begin{array}{ccc}1& 1& 2\\\\ 2& 4& -3\\\\ 3& 6& -5\\end{array}\\right][\/latex]<\/p>\n

4.\u00a0[latex]X=\\left[\\begin{array}{c}4\\\\ 38\\\\ 58\\end{array}\\right][\/latex]<\/p>\n

Solutions to Odd-Numbered Exercises<\/h2>\n

1.\u00a0If [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.<\/p>\n

3.\u00a0No, 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.<\/p>\n

5.\u00a0Yes. 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].<\/p>\n

7.\u00a0[latex]AB=BA=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]=I[\/latex]<\/p>\n

9.\u00a0[latex]AB=BA=\\left[\\begin{array}{cc}1& 0\\\\ 0& 1\\end{array}\\right]=I[\/latex]<\/p>\n

11.\u00a0[latex]AB=BA=\\left[\\begin{array}{ccc}1& 0& 0\\\\ 0& 1& 0\\\\ 0& 0& 1\\end{array}\\right]=I[\/latex]<\/p>\n

13.\u00a0[latex]\\frac{1}{29}\\left[\\begin{array}{cc}9& 2\\\\ -1& 3\\end{array}\\right][\/latex]<\/p>\n

15.\u00a0[latex]\\frac{1}{69}\\left[\\begin{array}{cc}-2& 7\\\\ 9& 3\\end{array}\\right][\/latex]<\/p>\n

17.\u00a0There is no inverse<\/p>\n

19.\u00a0[latex]\\frac{4}{7}\\left[\\begin{array}{cc}0.5& 1.5\\\\ 1& -0.5\\end{array}\\right][\/latex]<\/p>\n

21.\u00a0[latex]\\frac{1}{17}\\left[\\begin{array}{ccc}-5& 5& -3\\\\ 20& -3& 12\\\\ 1& -1& 4\\end{array}\\right][\/latex]<\/p>\n

23.\u00a0[latex]\\frac{1}{209}\\left[\\begin{array}{ccc}47& -57& 69\\\\ 10& 19& -12\\\\ -24& 38& -13\\end{array}\\right][\/latex]<\/p>\n

25.\u00a0[latex]\\left[\\begin{array}{ccc}18& 60& -168\\\\ -56& -140& 448\\\\ 40& 80& -280\\end{array}\\right][\/latex]<\/p>\n

27.\u00a0[latex]\\left(-5,6\\right)[\/latex]<\/p>\n

29.\u00a0[latex]\\left(2,0\\right)[\/latex]<\/p>\n

31.\u00a0[latex]\\left(\\frac{1}{3},-\\frac{5}{2}\\right)[\/latex]<\/p>\n

33.\u00a0[latex]\\left(-\\frac{2}{3},-\\frac{11}{6}\\right)[\/latex]<\/p>\n

35.\u00a0[latex]\\left(7,\\frac{1}{2},\\frac{1}{5}\\right)[\/latex]<\/p>\n

37.\u00a0[latex]\\left(5,0,-1\\right)[\/latex]<\/p>\n

39.\u00a0[latex]\\frac{1}{34}\\left(-35,-97,-154\\right)[\/latex]<\/p>\n

41.\u00a0[latex]\\frac{1}{690}\\left(65,-1136,-229\\right)[\/latex]<\/p>\n

43.\u00a0[latex]\\left(-\\frac{37}{30},\\frac{8}{15}\\right)[\/latex]<\/p>\n

45.\u00a0[latex]\\left(\\frac{10}{123},-1,\\frac{2}{5}\\right)[\/latex]<\/p>\n

47.\u00a0[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]<\/p>\n

49.\u00a0[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]<\/p>\n

51.\u00a0[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]<\/p>\n

53.\u00a0Infinite solutions.<\/p>\n

55.\u00a050% oranges, 25% bananas, 20% apples<\/p>\n

57.\u00a010 straw hats, 50 beanies, 40 cowboy hats<\/p>\n

59.\u00a0Tom ate 6, Joe ate 3, and Albert ate 3.<\/p>\n

61.\u00a0124 oranges, 10 lemons, 8 pomegranates<\/p>\n\n\t\t\t

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