{"id":61502,"date":"2020-09-02T15:45:29","date_gmt":"2020-09-02T15:45:29","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=61502"},"modified":"2026-03-26T09:59:02","modified_gmt":"2026-03-26T14:59:02","slug":"matrices-elementary-row-operations","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/matrices-elementary-row-operations\/","title":{"rendered":"Matrices: Elementary Row Operations"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_KeRkM805y4o\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_KeRkM805y4o\" data-source-videoID=\"KeRkM805y4o\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Matrices: Elementary Row Operations Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Matrices: Elementary Row Operations\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_KeRkM805y4o:hover {cursor:pointer;} img#videoThumbnailImage_KeRkM805y4o {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1684-matrices-elementary-row-operations-1-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_KeRkM805y4o\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_KeRkM805y4o\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Matrices: Elementary Row Operations\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_KeRkM805y4o\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_KeRkM805y4o\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_KeRkM805y4o\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction xs2_Function() {\n  var x = document.getElementById(\"xs2\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"xs2_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"xs2\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Elementary_Row_Operations\" class=\"smooth-scroll\">Elementary Row Operations<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Row_Echelon_Form\" class=\"smooth-scroll\">Row Echelon Form<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Reduced_Row_Echelon_Form\" class=\"smooth-scroll\">Reduced Row Echelon Form<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Inverse_Matrices\" class=\"smooth-scroll\">Inverse Matrices<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Matrix_Practice_Questions\" class=\"smooth-scroll\">Matrix Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video about <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/matrices-data-systems\/\">matrix<\/a> manipulation! <\/p>\n<p>In this video, we\u2019re going to look at:<\/p>\n<ul>\n<li>Elementary row operations<\/li>\n<li>Row echelon form of a matrix<\/li>\n<li>Reduced row echelon form of a matrix<\/li>\n<li>Finding the inverse of a matrix<\/li>\n<\/ul>\n<h2><span id=\"Elementary_Row_Operations\" class=\"m-toc-anchor\"><\/span>Elementary Row Operations<\/h2>\n<p>\nMatrices of any dimension can be manipulated using three elementary row operations:<\/p>\n<ol>\n<li>Any two rows can be swapped.<\/li>\n<li>Any row can be multiplied by a non-zero scalar.<\/li>\n<li>A multiple of a row can be added to another row.<\/li>\n<\/ol>\n<p>The process of using row operations on a matrix is referred to as <strong>row reduction<\/strong>.<\/p>\n<h2><span id=\"Row_Echelon_Form\" class=\"m-toc-anchor\"><\/span>Row Echelon Form<\/h2>\n<p>\nThe ultimate goal of row reduction is to end up with an upper triangular matrix\u2014a matrix with all zero entries below the main diagonal.  When this is achieved, the matrix is said to be in <strong>row echelon form<\/strong>.  <\/p>\n<p>Specifically, to be in row echelon form:<\/p>\n<ol>\n<li>Rows with all zero entries must be at the bottom of the matrix<\/li>\n<li>The first non-zero element of a row must be to the right of the first non-zero element in the row above.<\/li>\n<\/ol>\n<p>Let\u2019s see some row reduction in action! Keep in mind that the series of row operations is not unique.  Also, the row echelon form of a matrix is not unique.  Different people might use different series of operations to end up with different row echelon forms of the same matrix.<\/p>\n<p>Here\u2019s [X]:<\/p>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n4 &#038; 7 &#038; 1 \\\\<br \/>\n3 &#038; 0 &#038; 9<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>My first operation will be to use the 2 to turn the 4 below it into a 0. I want to subtract 2 times row 1 from row 2 and replace row 2.<\/p>\n<p>In shorthand, this looks like \\(R_2 &#8211; 2R_1 = R_2\\).<\/p>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n4 &#8211; 2(2) &#038; 7 &#8211; 2(3) &#038; 1 &#8211; 2(5) \\\\<br \/>\n3 &#038; 0 &#038; 9<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Now, I want to use the 2 again to turn the 3 in the bottom left into a 0. To make this easier, I&#8217;ll use two steps:<\/p>\n<ol>\n<li>\\(2R_3 = R_3\\)<\/li>\n<\/ol>\n<div class=\"longmatrix-container\">\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n2(3) &#038; 2(0) &#038; 2(9)<br \/>\n\\end{bmatrix}<br \/>\n=<br \/>\n\\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n6 &#038; 0 &#038; 18<br \/>\n\\end{bmatrix}<br \/>\n\\]\n<\/div>\n<ol start=\"2\">\n<li>\\(R_3 &#8211; 3R_1 = R_3\\)<\/li>\n<\/ol>\n<p>\\[<br \/>\nX =<br \/>\n\\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n6 &#8211; 3(2) &#038; 0 &#8211; 3(3) &#038; 18 &#8211; 3(5)<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Now there&#8217;s only one more move to get this matrix to row echelon form. I need to use the 1 to make the negative 9 in row three into a 0. So \\(R_3 + 9R_2  = R_3\\).<\/p>\n<div class=\"longmatrix-container\">\n\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n0 + 9(0) &#038; -9 + 9(1) &#038; 3 + 9(-9)<br \/>\n\\end{bmatrix}<br \/>\n\\]\n<\/div>\n<p>This is one version of [X] in row echelon form:<\/p>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n0 &#038; 0 &#038; -78<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Another version might look like this, if we had decided, for example, to divide row 3 by row 2:<\/p>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n0 &#038; 0 &#038; -39<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>As you can see, we ended with an upper triangular matrix.  The first nonzero entry in each row is to the right of the nonzero entry in the row above. In other words, the 1 is to the right of the 2 and the negative 78 is to the right of the 1.<\/p>\n<h2><span id=\"Reduced_Row_Echelon_Form\" class=\"m-toc-anchor\"><\/span>Reduced Row Echelon Form<\/h2>\n<p>\nWe could go further and get [X] into reduced row echelon form. Reduced row echelon form can be obtained using different sequences of row operations, but this form is unique.<\/p>\n<p>A matrix has only one reduced row echelon form. In order to be in reduced row echelon form, a matrix must meet these conditions:<\/p>\n<ul>\n<li>Be in row echelon form.<\/li>\n<li>The leading entry in each nonzero row must be a 1.<\/li>\n<li>Wherever there is a leading 1, the entries in the rest of the column must be zeros.<\/li>\n<\/ul>\n<p>Let\u2019s pick up with [X] where we left off:<\/p>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n0 &#038; 0 &#038; -78<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>First, I need to turn that 2 into a 1: \\(\\frac{1}{2}R_1 = R_1\\).<\/p>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n\\frac{1}{2}(2) &#038; \\frac{1}{2}(3) &#038; \\frac{1}{2}(5) \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n0 &#038; 0 &#038; -78<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>The first column looks good. Now to column 2. I need to use the 1 to turn the three halves above it into a 0: \\(R_1 &#8211; \\frac{3}{2}R_2 = R_1\\).<\/p>\n<div class=\"longmatrix-container\">\n\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n1 &#8211; \\frac{3}{2}(0) &#038; \\frac{3}{2} &#8211; \\frac{3}{2}(1) &#038; \\frac{5}{2} &#8211; \\frac{3}{2}(-9) \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n0 &#038; 0 &#038; -78<br \/>\n\\end{bmatrix}<br \/>\n\\]\n<\/div>\n<p>Now columns 1 and 2 look good. Time to turn that -78 into a 1: \\(\\frac{-1}{78}R_3 = R_3\\).<\/p>\n<div class=\"longmatrix-container\">\n\\[<br \/>\nX = \\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 16 \\\\<br \/>\n0 &#038; 1 &#038; -9 \\\\<br \/>\n-\\frac{1}{78}(0) &#038; -\\frac{1}{78}(0) &#038; -\\frac{1}{78}(-78)<br \/>\n\\end{bmatrix}<br \/>\n\\]\n<\/div>\n<p>Row 3 looks good. All that\u2019s left is to use row 3 to change the 16 and -9 to zeroes: \\(-16R_3 + R_1 = R_1\\) and \\(R_2 + 9R_3 = R_2\\).<\/p>\n<div class=\"longmatrix-container\">\n\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n-16(0) + 1 &#038; -16(0) + 0 &#038; -16(1) + 16 \\\\<br \/>\n0 + 9(0) &#038; 1 + 9(0) &#038; -9 + 9(1) \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]\n<\/div>\n<p>\\[<br \/>\nX =<br \/>\n \\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 1 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>In this case, the reduced row echelon form of the matrix is \\(I_3\\). With square matrices, this will happen frequently, but it doesn\u2019t need to happen.<\/p>\n<p>These matrices are in reduced row echelon form as well:<\/p>\n<p>\\begin{bmatrix}<br \/>\n1 \\\\<br \/>\n0 \\\\<br \/>\n0<br \/>\n\\end{bmatrix}<\/p>\n<p>\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 2 \\\\<br \/>\n0 &#038; 1 &#038; 3<br \/>\n\\end{bmatrix}<\/p>\n<p>\\begin{bmatrix}<br \/>\n1 &#038; 3 &#038; 0 &#038; 4 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; 6 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 0 &#038; 1 &#038; 1<br \/>\n\\end{bmatrix}<\/p>\n<h2><span id=\"Inverse_Matrices\" class=\"m-toc-anchor\"><\/span>Inverse Matrices<\/h2>\n<p>\nSome matrices have inverse matrices, denoted by the superscript -1.  Inverse matrices satisfy the equation \\(AA^{-1}=A^{-1}A=I\\).  <\/p>\n<p>For instance, here we have [B] and the [B<sup>-1<\/sup>]:<\/p>\n<p>\\[<br \/>\nB =<br \/>\n \\begin{bmatrix}<br \/>\n4 &#038; 3 \\\\<br \/>\n5 &#038; 4<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>\\[<br \/>\nB^{-1} =<br \/>\n \\begin{bmatrix}<br \/>\n4 &#038; -3 \\\\<br \/>\n-5 &#038; 4<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Let&#8217;s try [BB<sup>-1<\/sup>]. To do this, we do \\(4\\times 4 = 16\\), then \\(3 \\times -5 = -15\\). For our next element, we&#8217;ll do \\(4\\times -3=-12\\), then \\(3 \\times 4 = 12\\). Then, we&#8217;re going to do \\(5\\times 4 = 20\\), then \\(4 \\times -5 = -20\\). Then, we&#8217;ll do \\(5 \\times -3 =-15\\), then \\(4 \\times 4 = 16\\).<\/p>\n<p>\\[<br \/>\nBB^{-1} =<br \/>\n \\begin{bmatrix}<br \/>\n16 + (-15) &#038; (-12) + 12 \\\\<br \/>\n20 + (-20) &#038; (-15) + 16<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>If we simplify this, we get \\(16 + (-15) = 1\\), \\((-12) + 12 = 0\\), \\(20 + (-20) = 0\\), and \\((-15) + 16 = 1\\). <\/p>\n<p>\\[<br \/>\nBB^{-1} =<br \/>\n \\begin{bmatrix}<br \/>\n1 &#038; 0 \\\\<br \/>\n0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>This is our identity matrix for a 2\u00d72 matrix. If you go ahead and multiply [B<sup-1<\/sup> times [B], you&#8217;ll get the same thing.<\/p>\n<p>In order for a matrix to have an inverse, or to be <strong>invertible<\/strong>, two conditions must be met:<\/p>\n<ol>\n<li>The matrix must be a square matrix<\/li>\n<li>The determinant of the matrix must not be 0.<\/li>\n<\/ol>\n<p>Let\u2019s work with a 3\u00d73 matrix:<\/p>\n<p>\\[<br \/>\nY =<br \/>\n \\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; -2 \\\\<br \/>\n-4 &#038; 4 &#038; 1 \\\\<br \/>\n3 &#038; 0 &#038; -1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Is [Y] a square matrix? Check! Is the determinant of [Y] not equal to 0? Let\u2019s see.<\/p>\n<div style=\"display: flex; justify-content: center;\">\n<div style=\"text-align: left;\">\\(\\text{det(Y)}=2(-4)-3(1)+(-2)(-12)\\)<br \/>\n\\(\\phantom{\\text{det(Y)}}=-8-3+24\\)<br \/>\n\\(\\phantom{\\text{det(Y)}}=13\\)<\/div>\n<\/div>\n<p>\n&nbsp;<br \/>\nYes, the determinant of [Y] does not equal 0. So now let\u2019s figure out what the inverse of [Y] is.<\/p>\n<p>Here are the two steps to finding the inverse of a matrix:<\/p>\n<ol>\n<li>Augment the matrix with the identity matrix of the appropriate size.<br \/>\n(Y augment I looks like this: [Y|I])<\/li>\n<li>Row reduce the augmented matrix. When it is in reduced row echelon form, the structure will be [I|Y<sup>-1<\/sup>] and the inverse can be identified.<\/li>\n<\/ol>\n<p>Here&#8217;s what that process looks like:<\/p>\n<p>\\begin{bmatrix}<br \/>\n2 &#038; 3 &#038; -2 &#038; 1 &#038; 0 &#038; 0 \\\\<br \/>\n-4 &#038; 4 &#038; 1 &#038; 0 &#038; 1 &#038; 0 \\\\<br \/>\n3 &#038; 0 &#038; -1 &#038; 0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<\/p>\n<p>First, let&#8217;s turn the 2 into a 1: \\(\\frac{1}{2}R_1 = R_1\\).<\/p>\n<p>Next, let&#8217;s get zeroes below the 1 in column 1: \\(R_2 + 4R_1 = R_2\\) and \\(R_3 &#8211; 3R_1 = R_3\\).<\/p>\n<p>Now we need to make the 10 in column 2 into a 1: \\(\\frac{1}{10}R_2 = R_2\\).<\/p>\n<p>The rest of the entries in column 2 need to be 0: \\(R_1 &#8211; \\frac{3}{2}R_2 = R_1\\) and \\(R_3 + \\frac{9}{2}R_2 = R_3\\).<\/p>\n<p>Now the 13 over 20 in column 3 needs to change to 1: \\(\\frac{20}{13}R_3 = R_3\\).<\/p>\n<p>Lastly, the two remaining entries in column 3 need to be 0: \\(R_1 + \\frac{11}{20}R_3 = R_1\\) and \\(R_2 + \\frac{3}{10}R_3 = R_2\\).<\/p>\n<p>Now we have an augmented matrix in the form \\([I|Y^{-1}\\)]. <\/p>\n<p>\\[<br \/>\nI | Y^{-1} =<br \/>\n\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; -\\frac{4}{13} &#038; \\frac{3}{13} &#038; \\frac{11}{13} \\\\<br \/>\n0 &#038; 1 &#038; 0 &#038; -\\frac{1}{13} &#038; \\frac{4}{13} &#038; \\frac{6}{13} \\\\<br \/>\n0 &#038; 0 &#038; 1 &#038; -\\frac{12}{13} &#038; \\frac{9}{13} &#038; \\frac{20}{13}<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Matrix_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Matrix Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich step would change the 5 in \\([X]\\) into a zero?<\/p>\n<p style=\"text-align: center; margin-bottom: 1em\">\\(X=\\begin{bmatrix}<br \/>\n1 &#038; 2 &#038; 1\\\\<br \/>\n5 &#038; 3 &#038;0 \\\\<br \/>\n9 &#038; 8 &#038; 4<br \/>\n\\end{bmatrix}<br \/>\n\\)<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(R_1-5R_2=R_2\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">\\(R_2-5R_1=R_2\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(R_3-5R_2=R_2\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(R_2-5R_3=R_2\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>We can use the values in the matrices in place of the \\(R\\)&#8216;s the same way we substitute variables to see which one would result in 5 changing to 0.<\/p>\n<p>We can see when substituting the values in the equation \\(R_2-5R_1=R_2\\), we get \\(5-5(1)=0\\).<\/p>\n<p>Therefore, \\(R_2-5R_1=R_2\\) would change the 5 to a 0.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich matrix is in row echelon form?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">\\(\\begin{bmatrix}\r\n2 &#038; 6 &#038; -9\\\\ \r\n0 &#038; 5 &#038; 7\\\\ \r\n0 &#038; 0 &#038; 18\r\n\\end{bmatrix}\r\n\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(\\begin{bmatrix}\r\n1 &#038; 0 &#038; 0\\\\ \r\n 7&#038; 9 &#038; 0\\\\ \r\n4 &#038; -3 &#038; -1\r\n\\end{bmatrix}\r\n\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(\\begin{bmatrix}\r\n0 &#038; 0 &#038; 0\\\\ \r\n0 &#038; -1 &#038; 20\\\\ \r\n-4 &#038;5  &#038; 6\r\n\\end{bmatrix}\r\n\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\(\\begin{bmatrix}\r\n6 &#038; 4 &#038; 2\\\\ \r\n9 &#038; -8 &#038; 0\\\\ \r\n0 &#038; 0 &#038; 0\r\n\\end{bmatrix}\r\n\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>For a matrix to be in row echelon form, it must have all zero entries below the main diagonal line. The values 2, 5, and 18 make up the diagonal line of the matrix in Choice A and the entries below this line are zeros.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich matrix is in reduced row echelon form?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(\\begin{bmatrix}\r\n0 &#038; 1 &#038; 1\\\\ \r\n0 &#038; 1 &#038; 0\\\\ \r\n1 &#038; 1 &#038; 0\r\n\\end{bmatrix}\r\n\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(\\begin{bmatrix}\r\n1 &#038; 1 &#038; 1\\\\ \r\n0 &#038; 0 &#038; 0\\\\ \r\n0 &#038; 1 &#038; 0\r\n\\end{bmatrix}\r\n\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-3\">\\(\\begin{bmatrix}\r\n1 &#038; 0 &#038; 0\\\\ \r\n0 &#038; 1 &#038;0 \\\\ \r\n0 &#038; 0 &#038; 1\r\n\\end{bmatrix}\r\n\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(\\begin{bmatrix}\r\n1 &#038; 0 &#038; 0\\\\ \r\n0 &#038; 0 &#038; 1\\\\ \r\n1 &#038; 0 &#038; 1\r\n\\end{bmatrix}\r\n\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The definition of a matrix in reduced row echelon form is that it must be in row echelon form, the leading entry for each row that is nonzero must be a one, and where there is a leading one, the remaining entries must be zeros.<\/p>\n<p>The only matrix that meets these conditions is Choice C.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich is a true statement about a matrix and its inverse?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(A-A^{-1}=A^{-1}-A=I\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\(A+A^{-1}=A^{-1}+A=I\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(A\\div A^{-1}=A^{-1}\\div A=I\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-4\">\\(A\\times A^{-1}=A^{-1}\\times A=I\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The relationship between a matrix and its inverse states that if the matrix is multiplied by its inverse and vice versa, the resulting matrix is the identity matrix, which is what the statement \\((A\\times A^{-1}=A^{-1}\\times A=I)\\) is saying.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich statement is true about a matrix that is invertible?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">A matrix is invertible if it is a square matrix, and its determinant is zero.<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-2\">A matrix is invertible if it is a square matrix, and its determinant is not a zero.<\/div><div class=\"PQ\"  id=\"PQ-5-3\">A matrix is invertible if it is not a square matrix, and its determinant is zero.<\/div><div class=\"PQ\"  id=\"PQ-5-4\">A matrix is invertible if it is not a square matrix, and its determinant is not a zero.<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>A matrix is invertible if it is a square matrix, and its determinant is not a zero.<\/p>\n<p>For a matrix to be invertible it must meet two conditions: the matrix must be a square matrix, and the determinant cannot be a zero.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-ii\/\">Return to Algebra II Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra II 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