{"id":2485,"date":"2024-07-30T01:00:28","date_gmt":"2024-07-30T01:00:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=2485"},"modified":"2024-11-25T20:02:21","modified_gmt":"2024-11-25T20:02:21","slug":"solving-system-of-equations-using-matrices-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/solving-system-of-equations-using-matrices-apply-it-1\/","title":{"raw":"Solving System of Equations using Matrices: Apply It 1","rendered":"Solving System of Equations using Matrices: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Write the augmented matrix for a system of equations<\/li>\r\n \t<li>Write the system of equations from an augmented matrix<\/li>\r\n \t<li>Perform row operations on a matrix<\/li>\r\n \t<li>Use matrix operations and row reductions to find solutions to systems of linear equations<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Applications of Systems of Equations<\/h2>\r\nNow we will turn to the applications for which systems of equations are used. In the next example we determine how much money was invested at two different rates given the sum of the interest earned by both accounts.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Setting up a system of equations in the following examples uses the same ideas you have used before to write a system of linear equations to model a situation. The only difference is that now you'll use matrices and Gaussian elimination to solve the system.<\/section><section class=\"textbox example\" aria-label=\"Example\">Carolyn invests a total of [latex]$12,000[\/latex] in two municipal bonds, one paying [latex]10.5\\%[\/latex] interest and the other paying [latex]12\\%[\/latex] interest. The annual interest earned on the two investments last year was [latex]$1,335[\/latex]. How much was invested at each rate?[reveal-answer q=\"591342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"591342\"]We have a system of two equations in two variables. Let [latex]x=[\/latex] the amount invested at 10.5% interest, and [latex]y=[\/latex] the amount invested at 12% interest.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y=12,000\\hfill \\\\ 0.105x+0.12y=1,335\\hfill \\end{array}[\/latex]<\/p>\r\nAs a matrix, we have\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 12,000\\\\ \\hfill 0.105&amp; \\hfill 0.12&amp; \\hfill 1,335\\\\ \\end{array}\\right][\/latex]<\/p>\r\nMultiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 12,000\\\\ \\hfill 0&amp; \\hfill 0.015&amp; \\hfill 75\\\\ \\end{array}\\right][\/latex]<\/p>\r\nThen,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0.015y=75\\hfill \\\\ \\text{ }y=5,000\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]12,000 - 5,000=7,000[\/latex].\r\n\r\nThus, [latex]$5,000 [\/latex]was invested at [latex]12\\%[\/latex] interest and [latex]$7,000 [\/latex]at [latex]10.5\\%[\/latex] interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24852[\/ohm2_question]<\/section><section class=\"textbox example\" aria-label=\"Example\">Ava invests a total of [latex]$10,000[\/latex] in three accounts, one paying [latex]5\\%[\/latex] interest, another paying [latex]8\\%[\/latex] interest, and the third paying [latex]9\\%[\/latex] interest. The annual interest earned on the three investments last year was [latex]$770[\/latex]. The amount invested at [latex]9\\%[\/latex] was twice the amount invested at [latex]5\\%[\/latex]. How much was invested at each rate?[reveal-answer q=\"357628\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"357628\"]We have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at [latex]5\\%[\/latex] interest, let [latex]y[\/latex] be the amount invested at [latex]8\\%[\/latex] interest, and let [latex]z[\/latex] be the amount invested at [latex]9\\%[\/latex] interest. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y+z=10,000\\hfill \\\\ 0.05x+0.08y+0.09z=770\\hfill \\\\ \\text{ }2x-z=0\\hfill \\end{array}[\/latex]<\/p>\r\nAs a matrix, we have\r\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0.05&amp; \\hfill 0.08&amp; \\hfill 0.09&amp; \\hfill 770\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -1&amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\nNow, we perform Gaussian elimination to achieve row-echelon form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ -0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 0.03&amp; \\hfill 0.04&amp; \\hfill 270\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -1&amp; \\hfill 0\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ -2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 0.03&amp; \\hfill 0.04&amp; \\hfill 270\\\\ \\hfill 0&amp; -2&amp; \\hfill -3&amp; \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ \\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 9,000\\\\ \\hfill 0&amp; -2&amp; \\hfill -3&amp; \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ 2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1&amp; \\hfill 10,000\\\\ \\hfill 0&amp; \\hfill 1&amp; \\hfill \\frac{4}{3}&amp; \\hfill 9,000\\\\ \\hfill 0&amp; 0&amp; \\hfill -\\frac{1}{3}&amp; \\hfill -2,000 \\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\nThe third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].\r\n\r\nThe second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex].\r\n\r\nSubstituting [latex]z=6,000[\/latex], we get\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill y+\\frac{4}{3}\\left(6,000\\right)=9,000\\\\ \\hfill y+8,000=9,000\\\\ \\hfill y=1,000\\end{array}[\/latex]<\/p>\r\nThe first row tells us [latex]x+y+z=10,000[\/latex]. Substituting [latex]y=1,000[\/latex] and [latex]z=6,000[\/latex], we get\r\n[latex]\\begin{array}{l}x+1,000+6,000=10,000\\hfill \\\\ \\text{ }x=3,000\\text{ }\\hfill \\end{array}[\/latex]\r\n\r\nThe answer is [latex]$3,000[\/latex] invested at [latex]5\\%[\/latex] interest, [latex]$1,000[\/latex] invested at [latex]8\\%[\/latex], and [latex]$6,000[\/latex] invested at [latex]9\\%[\/latex] interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24853[\/ohm2_question]<\/section><section class=\"textbox example\" aria-label=\"Example\">A small shoe company took out a loan of [latex]$1,500,000[\/latex] to expand their inventory. Part of the money was borrowed at [latex]7\\%[\/latex], part was borrowed at [latex]8\\%[\/latex], and part was borrowed at [latex]10\\%[\/latex]. The amount borrowed at [latex]10\\%[\/latex] was four times the amount borrowed at [latex]7\\%[\/latex], and the annual interest on all three loans was [latex]$130,500[\/latex]. Use matrices to find the amount borrowed at each rate.[reveal-answer q=\"65592\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65592\"][latex]$150,000[\/latex] at [latex]7\\%[\/latex], [latex]$750,000[\/latex] at [latex]8\\%[\/latex], [latex]$600,000[\/latex] at [latex]10\\%[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24854[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Write the augmented matrix for a system of equations<\/li>\n<li>Write the system of equations from an augmented matrix<\/li>\n<li>Perform row operations on a matrix<\/li>\n<li>Use matrix operations and row reductions to find solutions to systems of linear equations<\/li>\n<\/ul>\n<\/section>\n<h2>Applications of Systems of Equations<\/h2>\n<p>Now we will turn to the applications for which systems of equations are used. In the next example we determine how much money was invested at two different rates given the sum of the interest earned by both accounts.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Setting up a system of equations in the following examples uses the same ideas you have used before to write a system of linear equations to model a situation. The only difference is that now you&#8217;ll use matrices and Gaussian elimination to solve the system.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Carolyn invests a total of [latex]$12,000[\/latex] in two municipal bonds, one paying [latex]10.5\\%[\/latex] interest and the other paying [latex]12\\%[\/latex] interest. The annual interest earned on the two investments last year was [latex]$1,335[\/latex]. How much was invested at each rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q591342\">Show Solution<\/button><\/p>\n<div id=\"q591342\" class=\"hidden-answer\" style=\"display: none\">We have a system of two equations in two variables. Let [latex]x=[\/latex] the amount invested at 10.5% interest, and [latex]y=[\/latex] the amount invested at 12% interest.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y=12,000\\hfill \\\\ 0.105x+0.12y=1,335\\hfill \\end{array}[\/latex]<\/p>\n<p>As a matrix, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 1& \\hfill 12,000\\\\ \\hfill 0.105& \\hfill 0.12& \\hfill 1,335\\\\ \\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{cc|c}\\hfill 1& \\hfill 1& \\hfill 12,000\\\\ \\hfill 0& \\hfill 0.015& \\hfill 75\\\\ \\end{array}\\right][\/latex]<\/p>\n<p>Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0.015y=75\\hfill \\\\ \\text{ }y=5,000\\hfill \\end{array}[\/latex]<\/p>\n<p>So [latex]12,000 - 5,000=7,000[\/latex].<\/p>\n<p>Thus, [latex]$5,000[\/latex]was invested at [latex]12\\%[\/latex] interest and [latex]$7,000[\/latex]at [latex]10.5\\%[\/latex] interest.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24852\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24852&theme=lumen&iframe_resize_id=ohm24852&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Ava invests a total of [latex]$10,000[\/latex] in three accounts, one paying [latex]5\\%[\/latex] interest, another paying [latex]8\\%[\/latex] interest, and the third paying [latex]9\\%[\/latex] interest. The annual interest earned on the three investments last year was [latex]$770[\/latex]. The amount invested at [latex]9\\%[\/latex] was twice the amount invested at [latex]5\\%[\/latex]. How much was invested at each rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q357628\">Show Solution<\/button><\/p>\n<div id=\"q357628\" class=\"hidden-answer\" style=\"display: none\">We have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at [latex]5\\%[\/latex] interest, let [latex]y[\/latex] be the amount invested at [latex]8\\%[\/latex] interest, and let [latex]z[\/latex] be the amount invested at [latex]9\\%[\/latex] interest. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+y+z=10,000\\hfill \\\\ 0.05x+0.08y+0.09z=770\\hfill \\\\ \\text{ }2x-z=0\\hfill \\end{array}[\/latex]<\/p>\n<p>As a matrix, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0.05& \\hfill 0.08& \\hfill 0.09& \\hfill 770\\\\ \\hfill 2& \\hfill 0& \\hfill -1& \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p>Now, we perform Gaussian elimination to achieve row-echelon form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ -0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 0.03& \\hfill 0.04& \\hfill 270\\\\ \\hfill 2& \\hfill 0& \\hfill -1& \\hfill 0\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ -2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 0.03& \\hfill 0.04& \\hfill 270\\\\ \\hfill 0& -2& \\hfill -3& \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ \\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{4}{3}& \\hfill 9,000\\\\ \\hfill 0& -2& \\hfill -3& \\hfill -20,000 \\end{array}\\right]\\hfill \\\\ 2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 1& \\hfill 1& \\hfill 10,000\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{4}{3}& \\hfill 9,000\\\\ \\hfill 0& 0& \\hfill -\\frac{1}{3}& \\hfill -2,000 \\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\n<p>The third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].<\/p>\n<p>The second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex].<\/p>\n<p>Substituting [latex]z=6,000[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill y+\\frac{4}{3}\\left(6,000\\right)=9,000\\\\ \\hfill y+8,000=9,000\\\\ \\hfill y=1,000\\end{array}[\/latex]<\/p>\n<p>The first row tells us [latex]x+y+z=10,000[\/latex]. Substituting [latex]y=1,000[\/latex] and [latex]z=6,000[\/latex], we get<br \/>\n[latex]\\begin{array}{l}x+1,000+6,000=10,000\\hfill \\\\ \\text{ }x=3,000\\text{ }\\hfill \\end{array}[\/latex]<\/p>\n<p>The answer is [latex]$3,000[\/latex] invested at [latex]5\\%[\/latex] interest, [latex]$1,000[\/latex] invested at [latex]8\\%[\/latex], and [latex]$6,000[\/latex] invested at [latex]9\\%[\/latex] interest.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24853\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24853&theme=lumen&iframe_resize_id=ohm24853&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A small shoe company took out a loan of [latex]$1,500,000[\/latex] to expand their inventory. Part of the money was borrowed at [latex]7\\%[\/latex], part was borrowed at [latex]8\\%[\/latex], and part was borrowed at [latex]10\\%[\/latex]. The amount borrowed at [latex]10\\%[\/latex] was four times the amount borrowed at [latex]7\\%[\/latex], and the annual interest on all three loans was [latex]$130,500[\/latex]. Use matrices to find the amount borrowed at each rate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q65592\">Show Solution<\/button><\/p>\n<div id=\"q65592\" class=\"hidden-answer\" style=\"display: none\">[latex]$150,000[\/latex] at [latex]7\\%[\/latex], [latex]$750,000[\/latex] at [latex]8\\%[\/latex], [latex]$600,000[\/latex] at [latex]10\\%[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24854\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24854&theme=lumen&iframe_resize_id=ohm24854&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":327,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2485"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2485\/revisions"}],"predecessor-version":[{"id":6482,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2485\/revisions\/6482"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/327"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2485\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=2485"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2485"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=2485"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=2485"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}