Let’s begin by correcting and rewriting the system of equations properly. The given system seems to contain typographical errors in the constants. Specifically, \u201c212\u201d and \u201c42\u201d should likely be expressions involving the variable z<\/strong>, such as +21z<\/strong> and +4z<\/strong>, based on typical equation structure.<\/p>\n\n\n\n Assuming the corrected system of equations is:<\/p>\n\n\n\n 1.\u20033x + 27y + 21z = 45 [327214518\u22126021840]\\begin{bmatrix} 3 & 27 & 21 & 45 \\\\ 1 & 8 & -6 & 0 \\\\ 2 & 18 & 4 & 0 \\end{bmatrix}\u200b312\u200b27818\u200b21\u221264\u200b4500\u200b\u200b<\/p>\n\n\n\n To input and solve using Using the calculator or performing manually, the reduced matrix is:[10060100001\u22123]\\begin{bmatrix} 1 & 0 & 0 & 6 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & -3 \\end{bmatrix}\u200b100\u200b010\u200b001\u200b60\u22123\u200b\u200b<\/p>\n\n\n\n From the matrix, this means:<\/p>\n\n\n\n x = 6<\/strong><\/p>\n\n\n\n To solve a system of linear equations using matrices, we can represent the system in an augmented matrix format and then simplify it using the Reduced Row Echelon Form (RREF). This form is useful because it allows direct reading of the solution to the system.<\/p>\n\n\n\n In this problem, we began with a system of three linear equations involving three variables: x, y, and z. The goal was to find the values of these variables that satisfy all three equations simultaneously. By expressing the system as an augmented matrix, we transformed the equations into numerical form, placing the coefficients of x, y, and z along with the constants on the right-hand side.<\/p>\n\n\n\n The TI-83\/84 calculator has a built-in function called After entering the matrix and applying the This method is especially useful in systems with three or more variables, where substitution or elimination by hand becomes time-consuming. The calculator performs all the row operations efficiently, making the solution process faster and less prone to arithmetic errors.<\/p>\n\n\n\n Enter the following system as matrix [A] into your calculator and reduce the matrix using the rref() command (MATRIX – MATH B:rref() on the TI 83\/84): 3x + 27y + 212 = 45 x + 8y – 6z = 0 2x + 18y + 42 = 0 The reduced matrix supplied by the calculator using […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-39061","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39061","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=39061"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39061\/revisions"}],"predecessor-version":[{"id":39063,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39061\/revisions\/39063"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39061"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39061"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39061"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
2.\u2003x + 8y \u2212 6z = 0
3.\u20032x + 18y + 4z = 0<\/p>\n\n\n\n
\n\n\n\nStep 1: Write the augmented matrix [A]<\/h3>\n\n\n\n
\n\n\n\nStep 2: Use your TI-83\/84 Calculator<\/h3>\n\n\n\n
rref()<\/code>:<\/p>\n\n\n\n\n
2nd<\/code> then MATRIX<\/code>.<\/li>\n\n\n\nEDIT<\/code>, select [A], and enter a 3×4 matrix.<\/li>\n\n\n\n2nd<\/code> then QUIT<\/code>.<\/li>\n\n\n\nMATRIX<\/code>, choose MATH<\/code>, scroll to B:rref(<\/code>.<\/li>\n\n\n\nrref([A])<\/code> and press ENTER<\/code>.<\/li>\n<\/ol>\n\n\n\n
\n\n\n\nStep 3: The reduced row echelon form (RREF)<\/h3>\n\n\n\n
\n\n\n\nStep 4: Interpret the solution<\/h3>\n\n\n\n
\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
rref()<\/code>, which stands for Reduced Row Echelon Form. This function applies the Gaussian elimination method followed by back-substitution to simplify the matrix so that the leading coefficient of each row is 1 and all other entries in the pivot column are zero.<\/p>\n\n\n\nrref()<\/code> function, the output matrix clearly showed the identity matrix on the left and the solution column on the right. This indicated that the system has a unique solution: x = 6, y = 0, and z = -3. From this matrix form, it is easy to interpret the result, since each row corresponds to a solved equation: x = 6, y = 0, and z = -3.<\/p>\n\n\n\n![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1117.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"