{"id":2480,"date":"2024-07-30T00:43:56","date_gmt":"2024-07-30T00:43:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2480"},"modified":"2025-01-29T22:26:41","modified_gmt":"2025-01-29T22:26:41","slug":"solving-system-of-equations-using-matrices-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/solving-system-of-equations-using-matrices-learn-it-4\/","title":{"raw":"Solving System of Equations using Matrices: Learn It 4","rendered":"Solving System of Equations using Matrices: Learn It 4"},"content":{"raw":"

Solving a System of Linear Equations Using Matrices<\/h2>\r\nWe have seen how to write a system of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.\r\n\r\n
Solve the system of linear equations using matrices.\r\n

[latex]\\begin{array}{c}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ x-y+z=8\\hfill \\end{array}\\\\ 2x+3y-z=-2\\\\ 3x - 2y - 9z=9\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"779369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"779369\"]\r\n\r\nFirst, we write the augmented matrix.\r\n

[latex]\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 2& \\hfill 3& \\hfill -1& \\hfill -2\\\\ \\hfill 3& \\hfill -2& \\hfill -9& \\hfill 9\\end{array}\\right][\/latex]<\/p>\r\n

Next, we perform row operations to obtain row-echelon form.<\/p>\r\n

[latex]\\begin{array}{rrrrr}\\hfill -2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 5& \\hfill -3& \\hfill -18\\\\ \\hfill 3& \\hfill -2& \\hfill -9& \\hfill 9\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 5& \\hfill -3& \\hfill -18\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\end{array}\\right]\\end{array}[\/latex]<\/p>\r\nThe easiest way to obtain a 1 in row 2 of column 1 is to interchange [latex]{R}_{2}[\/latex] and [latex]{R}_{3}[\/latex].\r\n

[latex]\\text{Interchange }{R}_{2}\\text{ and }{R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\\\ \\hfill 0& \\hfill 5& \\hfill -3& \\hfill -18\\end{array}\\right][\/latex]<\/p>\r\nThen\r\n

[latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill -5{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\\\ \\hfill 0& \\hfill 0& \\hfill 57& \\hfill 57\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -\\frac{1}{57}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill -1& \\hfill 1& \\hfill 8\\\\ \\hfill 0& \\hfill 1& \\hfill -12& \\hfill -15\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/p>\r\nThe last matrix represents the equivalent system.\r\n

[latex]\\begin{array}{l}\\text{ }x-y+z=8\\hfill \\\\ \\text{ }y - 12z=-15\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/p>\r\nUsing back-substitution, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>Recall that there are three possible outcomes for solutions to linear systems. \u00a0In the previous example, the solution\u00a0[latex]\\left(4,-3,1\\right)[\/latex] represents a point in three dimensional space. This point represents the intersection of three planes. \u00a0In the next example, we solve a system using row operations and find that it represents a dependent system. \u00a0A dependent system in 3 dimensions can be represented by two planes that are identical, much like in 2 dimensions where a dependent system represents two lines that are identical.\r\n\r\n

Solve the following system of linear equations using Gaussian Elimination.\r\n

[latex]\\begin{array}{r}\\hfill -x - 2y+z=-1\\\\ \\hfill 2x+3y=2\\\\ \\hfill y - 2z=0\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"664612\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664612\"]\r\n\r\nWrite the augmented matrix.\r\n

[latex]\\left[\\begin{array}{ccc|c}\\hfill -1& \\hfill -2& \\hfill 1& \\hfill -1\\\\ \\hfill 2& \\hfill 3& \\hfill 0& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\nFirst, multiply row 1 by [latex]-1[\/latex] to get a 1 in row 1, column 1. Then, perform row operations<\/strong> to obtain row-echelon form.\r\n

[latex]-{R}_{1}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 2& \\hfill 3& \\hfill 0& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n

[latex]{R}_{2}\\leftrightarrow {R}_{3}\\to \\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 2& \\hfill 3& \\hfill 0& \\hfill 2\\end{array}\\right][\/latex]<\/p>\r\n

[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 1\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill -1& \\hfill 2& \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n

[latex]{R}_{2}+{R}_{3}={R}_{3}\\to\\left[\\begin{array}{ccc|c}\\hfill 1& \\hfill 2& \\hfill -1& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -2& \\hfill 1\\\\ \\hfill 0& \\hfill 0& \\hfill 0& \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\nThe last matrix represents the following system.\r\n

[latex]\\begin{array}{l}\\text{ }x+2y-z=1\\hfill \\\\ \\text{ }y - 2z=0\\hfill \\\\ \\text{ }0=0\\hfill \\end{array}[\/latex]<\/p>\r\nWe see by the identity [latex]0=0[\/latex] that this is a dependent system with an infinite number of solutions. We then find the generic solution.\r\n\r\nFirst, solve the second equation for [latex]y[\/latex]:\r\n

[latex]\\begin{align} y - 2z &= 0 \\\\ y &= 2z \\end{align}[\/latex]<\/p>\r\nNext, substitute [latex]y = 2z[\/latex] into the first equation:\r\n

[latex]\\begin{align} x + 2(2z) - z &= 1 \\\\ x + 4z - z &= 1 \\\\ x + 3z &= 1 \\\\ x &= 1 - 3z \\end{align}[\/latex]<\/p>\r\nThe generic solution is [latex](1-3z,2z,z)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>Recall that when you solve a dependent system of linear equations in two variables using elimination or substitution, you can write the solution [latex](x,y)[\/latex] in terms of [latex]x[\/latex], because there are infinitely many [latex](x,y)[\/latex] pairs that will satisfy a dependent system of equations, and they all fall on the line [latex](x, mx+b)[\/latex]. Now that you are working in three dimensions, the solution will represent a plane, so you would write it in the general form [latex](x, m_{1}x+b_{1}, m_{2}x+b_{2})[\/latex]. However, in three dimensions, it is often convenient to express the solution in terms of the single variable [latex]z[\/latex].\r\n\r\n

[ohm2_question hide_question_numbers=1]24850[\/ohm2_question]<\/section>
[ohm2_question hide_question_numbers=1]24851[\/ohm2_question]<\/section>
How To: Given a system of equations, solve with matrices using a calculator<\/strong>\r\n
    \r\n \t
  1. Save the augmented matrix as a matrix variable [latex]\\left[A\\right],\\left[B\\right],\\left[C\\right]\\text{,} \\dots [\/latex].<\/li>\r\n \t
  2. Use the ref(<\/strong> function in the calculator, calling up each matrix variable as needed.<\/li>\r\n<\/ol>\r\n<\/section>