{"id":1800,"date":"2015-11-12T18:30:44","date_gmt":"2015-11-12T18:30:44","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1800"},"modified":"2015-11-12T18:30:44","modified_gmt":"2015-11-12T18:30:44","slug":"solving-a-system-of-linear-equations-using-matrices","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/chapter\/solving-a-system-of-linear-equations-using-matrices\/","title":{"raw":"Solving a System of Linear Equations Using Matrices","rendered":"Solving a System of Linear Equations Using Matrices"},"content":{"raw":"

We have seen how to write a system of equations<\/strong> with an augmented matrix<\/strong>, and then how to use row operations and back-substitution to obtain row-echelon form<\/strong>. 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.\n<\/p>

\n

Example 6: Solving a System of Linear Equations Using Matrices<\/h3>\nSolve the system of linear equations using matrices.\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]<\/div>\n<\/div>\n
\n

Solution<\/h3>\nFirst, we write the augmented matrix.\n
[latex]\\left[\\begin{array}{rrr}\\hfill 1& \\hfill -1& \\hfill 1\\\\ \\hfill 2& \\hfill 3& \\hfill -1\\\\ \\hfill 3& \\hfill -2& \\hfill -9\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 8\\\\ \\hfill -2\\\\ \\hfill 9\\end{array}\\right][\/latex]<\/div>\nNext, we perform row operations to obtain row-echelon form.\n
[latex]\\begin{array}{rrrrr}\\hfill -2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 5& \\hfill & \\hfill -3& \\hfill \\\\ \\hfill 3& \\hfill & \\hfill -2& \\hfill & \\hfill -9& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -18\\\\ \\hfill & \\hfill 9\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 5& \\hfill & \\hfill -3& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -18\\\\ \\hfill & \\hfill -15\\end{array}\\right]\\end{array}[\/latex]<\/div>\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].\n
[latex]\\text{Interchange}{R}_{2}\\text{and}{R}_{3}\\to \\left[\\begin{array}{rrrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill & \\hfill 8\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill & \\hfill -15\\\\ \\hfill 0& \\hfill & \\hfill 5& \\hfill & \\hfill -3& \\hfill & \\hfill -18\\end{array}\\right][\/latex]<\/div>\nThen\n
[latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill -5{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 57& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -15\\\\ \\hfill & \\hfill 57\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -\\frac{1}{57}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -15\\\\ \\hfill & \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/div>\nThe last matrix represents the equivalent system.\n
[latex]\\begin{array}{l}\\text{ }x-y+z=8\\hfill \\\\ \\text{ }y - 12z=-15\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/div>\nUsing back-substitution, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].\n\n<\/div>\n
\n

Example 7: Solving a Dependent System of Linear Equations Using Matrices<\/h3>\nSolve the following system of linear equations using matrices.\n
[latex]\\begin{array}{r}\\hfill -x - 2y+z=-1\\\\ \\hfill 2x+3y=2\\\\ \\hfill y - 2z=0\\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Solution<\/h3>\nWrite the augmented matrix.\n
[latex]\\left[\\begin{array}{rrr}\\hfill -1& \\hfill -2& \\hfill 1\\\\ \\hfill 2& \\hfill 3& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill -2\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill -1\\\\ \\hfill 2\\\\ \\hfill 0\\end{array}\\right][\/latex]<\/div>\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.\n
[latex]-{R}_{1}\\to \\left[\\begin{array}{rrrrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1& \\hfill & \\hfill 1\\\\ \\hfill 2& \\hfill & \\hfill 3& \\hfill & \\hfill 0& \\hfill & \\hfill 2\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2& \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
[latex]{R}_{2}\\leftrightarrow {R}_{3}\\to \\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2\\\\ \\hfill 2& \\hfill & \\hfill 3& \\hfill & \\hfill 0\\end{array}\\text{ }|\\begin{array}{rr}\\hfill & \\hfill 1\\\\ \\hfill & \\hfill 0\\\\ \\hfill & \\hfill 2\\end{array}\\right][\/latex]<\/div>\n
[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -1& \\hfill & \\hfill 2& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 1\\\\ \\hfill & \\hfill 0\\\\ \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
[latex]{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 0& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 2\\\\ \\hfill & \\hfill 1\\\\ \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/div>\nThe last matrix represents the following system.\n
[latex]\\begin{array}{l}\\text{ }x+2y-z=1\\hfill \\\\ \\text{ }y - 2z=0\\hfill \\\\ \\text{ }0=0\\hfill \\end{array}[\/latex]<\/div>\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. By solving the second equation for [latex]y[\/latex] and substituting it into the first equation we can solve for [latex]z[\/latex] in terms of [latex]x[\/latex].\n
[latex]\\begin{array}{l}\\text{ }x+2y-z=1\\hfill \\\\ \\text{ }y=2z\\hfill \\\\ \\hfill \\\\ x+2\\left(2z\\right)-z=1\\hfill \\\\ \\text{ }x+3z=1\\hfill \\\\ \\text{ }z=\\frac{1-x}{3}\\hfill \\end{array}[\/latex]<\/div>\nNow we substitute the expression for [latex]z[\/latex] into the second equation to solve for [latex]y[\/latex] in terms of [latex]x[\/latex].\n
[latex]\\begin{array}{l}\\text{ }y - 2z=0\\hfill \\\\ \\text{ }z=\\frac{1-x}{3}\\hfill \\\\ \\hfill \\\\ y - 2\\left(\\frac{1-x}{3}\\right)=0\\hfill \\\\ \\text{ }y=\\frac{2 - 2x}{3}\\hfill \\end{array}[\/latex]<\/div>\nThe generic solution is [latex]\\left(x,\\frac{2 - 2x}{3},\\frac{1-x}{3}\\right)[\/latex].\n\n<\/div>\n
\n

Try It\u00a05<\/h3>\nSolve the system using matrices.\n
[latex]\\begin{array}{c}x+4y-z=4\\\\ 2x+5y+8z=15\\\\ x+3y - 3z=1\\end{array}[\/latex]<\/div>\n
Solution<\/a><\/div>\n<\/div>\n
\n

Q & A<\/h3>\n

Can any system of linear equations be solved by Gaussian elimination?<\/h3>\nYes, a system of linear equations of any size can be solved by Gaussian elimination.<\/em>\n\n<\/div>\n
\n

How To: Given a system of equations, solve with matrices using a calculator.<\/h3>\n
  1. Save the augmented matrix as a matrix variable [latex]\\left[A\\right],\\left[B\\right],\\left[C\\right]\\text{,} \\dots [\/latex].<\/li>\n\t
  2. Use the ref(<\/strong> function in the calculator, calling up each matrix variable as needed.<\/li>\n<\/ol><\/div>\n
    \n

    Example 8: Solving Systems of Equations with Matrices Using a Calculator<\/h3>\nSolve the system of equations.\n
    [latex]\\begin{array}{r}\\hfill 5x+3y+9z=-1\\\\ \\hfill -2x+3y-z=-2\\\\ \\hfill -x - 4y+5z=1\\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\nWrite the augmented matrix for the system of equations.\n
    [latex]\\left[\\begin{array}{rrr}\\hfill 5& \\hfill 3& \\hfill 9\\\\ \\hfill -2& \\hfill 3& \\hfill -1\\\\ \\hfill -1& \\hfill -4& \\hfill 5\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 5\\\\ \\hfill -2\\\\ \\hfill -1\\end{array}\\right][\/latex]<\/div>\nOn the matrix page of the calculator, enter the augmented matrix above as the matrix variable [latex]\\left[A\\right][\/latex].\n
    [latex]\\left[A\\right]=\\left[\\begin{array}{rrrrrrr}\\hfill 5& \\hfill & \\hfill 3& \\hfill & \\hfill 9& \\hfill & \\hfill -1\\\\ \\hfill -2& \\hfill & \\hfill 3& \\hfill & \\hfill -1& \\hfill & \\hfill -2\\\\ \\hfill -1& \\hfill & \\hfill -4& \\hfill & \\hfill 5& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\nUse the ref(<\/strong> function in the calculator, calling up the matrix variable [latex]\\left[A\\right][\/latex].\n
    [latex]\\text{ref}\\left(\\left[A\\right]\\right)[\/latex]<\/div>\nEvaluate.\n
    [latex]\\begin{array}{l}\\hfill \\\\ \\left[\\begin{array}{rrrr}\\hfill 1& \\hfill \\frac{3}{5}& \\hfill \\frac{9}{5}& \\hfill \\frac{1}{5}\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{13}{21}& \\hfill -\\frac{4}{7}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill -\\frac{24}{187}\\end{array}\\right]\\to \\begin{array}{l}x+\\frac{3}{5}y+\\frac{9}{5}z=-\\frac{1}{5}\\hfill \\\\ \\text{ }y+\\frac{13}{21}z=-\\frac{4}{7}\\hfill \\\\ \\text{ }z=-\\frac{24}{187}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\nUsing back-substitution, the solution is [latex]\\left(\\frac{61}{187},-\\frac{92}{187},-\\frac{24}{187}\\right)[\/latex].\n\n<\/div>\n
    \n

    Example 9: Applying 2 \u00d7 2 Matrices to Finance<\/h3>\nCarolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?\n\n<\/div>\n
    \n

    Solution<\/h3>\nWe 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.\n
    [latex]\\begin{array}{l}\\text{ }x+y=12,000\\hfill \\\\ 0.105x+0.12y=1,335\\hfill \\end{array}[\/latex]<\/div>\nAs a matrix, we have\n
    [latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0.105& \\hfill 0.12\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 1,335\\end{array}\\right][\/latex]<\/div>\nMultiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.\n
    [latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill 0.015\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 75\\end{array}\\right][\/latex]<\/div>\nThen,\n
    [latex]\\begin{array}{l}0.015y=75\\hfill \\\\ \\text{ }y=5,000\\hfill \\end{array}[\/latex]<\/div>\nSo [latex]12,000 - 5,000=7,000[\/latex].\n\nThus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.\n\n<\/div>\n
    \n

    Example 10: Applying 3 \u00d7 3 Matrices to Finance<\/h3>\nAva invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?\n\n<\/div>\n
    \n

    Solution<\/h3>\nWe have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at 5% interest, let [latex]y[\/latex] be the amount invested at 8% interest, and let [latex]z[\/latex] be the amount invested at 9% interest. Thus,\n
    [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]<\/div>\nAs a matrix, we have\n
    [latex]\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 1& \\hfill 1\\\\ \\hfill 0.05& \\hfill 0.08& \\hfill 0.09\\\\ \\hfill 2& \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 10,000\\\\ \\hfill 770\\\\ \\hfill 0\\end{array}\\right][\/latex]<\/div>\nNow, we perform Gaussian elimination to achieve row-echelon form.\n
    [latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ -0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0.03& \\hfill & \\hfill 0.04& \\hfill \\\\ \\hfill 2& \\hfill & \\hfill 0& \\hfill & \\hfill -1& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 270\\\\ \\hfill & \\hfill 0\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ -2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0.03& \\hfill & \\hfill 0.04& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -2& \\hfill & \\hfill -3& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 270\\\\ \\hfill & \\hfill -20,000\\end{array}\\right]\\hfill \\\\ \\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{rrrrrr}\\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill \\frac{4}{3}& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -2& \\hfill & \\hfill -3& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 9,000\\\\ \\hfill & \\hfill -20,000\\end{array}\\right]\\hfill \\\\ 2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill \\frac{4}{3}& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill -\\frac{1}{3}& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 9,000\\\\ \\hfill & \\hfill -2,000\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\nThe third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].\n\nThe second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex]. Substituting [latex]z=6,000[\/latex], we get\n
    [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]<\/div>\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\n
    [latex]\\begin{array}{l}x+1,000+6,000=10,000\\hfill \\\\ \\text{ }x=3,000\\text{ }\\hfill \\end{array}[\/latex]<\/div>\nThe answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.\n\n<\/div>\n
    \n

    Try It 6<\/h3>\nA small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.\n\nSolution<\/a>\n\n<\/div>","rendered":"

    We have seen how to write a system of equations<\/strong> with an augmented matrix<\/strong>, and then how to use row operations and back-substitution to obtain row-echelon form<\/strong>. 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.\n<\/p>\n

    \n

    Example 6: Solving a System of Linear Equations Using Matrices<\/h3>\n

    Solve the system of linear equations using matrices.<\/p>\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]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\n

    First, we write the augmented matrix.<\/p>\n

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

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

    [latex]\\begin{array}{rrrrr}\\hfill -2{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 5& \\hfill & \\hfill -3& \\hfill \\\\ \\hfill 3& \\hfill & \\hfill -2& \\hfill & \\hfill -9& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -18\\\\ \\hfill & \\hfill 9\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -3{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 5& \\hfill & \\hfill -3& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -18\\\\ \\hfill & \\hfill -15\\end{array}\\right]\\end{array}[\/latex]<\/div>\n

    The easiest way to obtain a 1 in row 2 of column 1 is to interchange [latex]{R}_{2}[\/latex] and [latex]{R}_{3}[\/latex].<\/p>\n

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

    Then<\/p>\n

    [latex]\\begin{array}{l}\\\\ \\begin{array}{rrrrr}\\hfill -5{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 57& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -15\\\\ \\hfill & \\hfill 57\\end{array}\\right]& \\hfill & \\hfill & \\hfill & \\hfill -\\frac{1}{57}{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill -1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -12& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 1& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 8\\\\ \\hfill & \\hfill -15\\\\ \\hfill & \\hfill 1\\end{array}\\right]\\end{array}\\end{array}[\/latex]<\/div>\n

    The last matrix represents the equivalent system.<\/p>\n

    [latex]\\begin{array}{l}\\text{ }x-y+z=8\\hfill \\\\ \\text{ }y - 12z=-15\\hfill \\\\ \\text{ }z=1\\hfill \\end{array}[\/latex]<\/div>\n

    Using back-substitution, we obtain the solution as [latex]\\left(4,-3,1\\right)[\/latex].<\/p>\n<\/div>\n

    \n

    Example 7: Solving a Dependent System of Linear Equations Using Matrices<\/h3>\n

    Solve the following system of linear equations using matrices.<\/p>\n

    [latex]\\begin{array}{r}\\hfill -x - 2y+z=-1\\\\ \\hfill 2x+3y=2\\\\ \\hfill y - 2z=0\\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\n

    Write the augmented matrix.<\/p>\n

    [latex]\\left[\\begin{array}{rrr}\\hfill -1& \\hfill -2& \\hfill 1\\\\ \\hfill 2& \\hfill 3& \\hfill 0\\\\ \\hfill 0& \\hfill 1& \\hfill -2\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill -1\\\\ \\hfill 2\\\\ \\hfill 0\\end{array}\\right][\/latex]<\/div>\n

    First, 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.<\/p>\n

    [latex]-{R}_{1}\\to \\left[\\begin{array}{rrrrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1& \\hfill & \\hfill 1\\\\ \\hfill 2& \\hfill & \\hfill 3& \\hfill & \\hfill 0& \\hfill & \\hfill 2\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2& \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
    [latex]{R}_{2}\\leftrightarrow {R}_{3}\\to \\left[\\begin{array}{rrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1\\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2\\\\ \\hfill 2& \\hfill & \\hfill 3& \\hfill & \\hfill 0\\end{array}\\text{ }|\\begin{array}{rr}\\hfill & \\hfill 1\\\\ \\hfill & \\hfill 0\\\\ \\hfill & \\hfill 2\\end{array}\\right][\/latex]<\/div>\n
    [latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -1& \\hfill & \\hfill 2& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 1\\\\ \\hfill & \\hfill 0\\\\ \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/div>\n
    [latex]{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 2& \\hfill & \\hfill -1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill -2& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill 0& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 2\\\\ \\hfill & \\hfill 1\\\\ \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/div>\n

    The last matrix represents the following system.<\/p>\n

    [latex]\\begin{array}{l}\\text{ }x+2y-z=1\\hfill \\\\ \\text{ }y - 2z=0\\hfill \\\\ \\text{ }0=0\\hfill \\end{array}[\/latex]<\/div>\n

    We 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. By solving the second equation for [latex]y[\/latex] and substituting it into the first equation we can solve for [latex]z[\/latex] in terms of [latex]x[\/latex].<\/p>\n

    [latex]\\begin{array}{l}\\text{ }x+2y-z=1\\hfill \\\\ \\text{ }y=2z\\hfill \\\\ \\hfill \\\\ x+2\\left(2z\\right)-z=1\\hfill \\\\ \\text{ }x+3z=1\\hfill \\\\ \\text{ }z=\\frac{1-x}{3}\\hfill \\end{array}[\/latex]<\/div>\n

    Now we substitute the expression for [latex]z[\/latex] into the second equation to solve for [latex]y[\/latex] in terms of [latex]x[\/latex].<\/p>\n

    [latex]\\begin{array}{l}\\text{ }y - 2z=0\\hfill \\\\ \\text{ }z=\\frac{1-x}{3}\\hfill \\\\ \\hfill \\\\ y - 2\\left(\\frac{1-x}{3}\\right)=0\\hfill \\\\ \\text{ }y=\\frac{2 - 2x}{3}\\hfill \\end{array}[\/latex]<\/div>\n

    The generic solution is [latex]\\left(x,\\frac{2 - 2x}{3},\\frac{1-x}{3}\\right)[\/latex].<\/p>\n<\/div>\n

    \n

    Try It\u00a05<\/h3>\n

    Solve the system using matrices.<\/p>\n

    [latex]\\begin{array}{c}x+4y-z=4\\\\ 2x+5y+8z=15\\\\ x+3y - 3z=1\\end{array}[\/latex]<\/div>\n
    Solution<\/a><\/div>\n<\/div>\n
    \n

    Q & A<\/h3>\n

    Can any system of linear equations be solved by Gaussian elimination?<\/h3>\n

    Yes, a system of linear equations of any size can be solved by Gaussian elimination.<\/em><\/p>\n<\/div>\n

    \n

    How To: Given a system of equations, solve with matrices using a calculator.<\/h3>\n
      \n
    1. Save the augmented matrix as a matrix variable [latex]\\left[A\\right],\\left[B\\right],\\left[C\\right]\\text{,} \\dots[\/latex].<\/li>\n
    2. Use the ref(<\/strong> function in the calculator, calling up each matrix variable as needed.<\/li>\n<\/ol>\n<\/div>\n
      \n

      Example 8: Solving Systems of Equations with Matrices Using a Calculator<\/h3>\n

      Solve the system of equations.<\/p>\n

      [latex]\\begin{array}{r}\\hfill 5x+3y+9z=-1\\\\ \\hfill -2x+3y-z=-2\\\\ \\hfill -x - 4y+5z=1\\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\n

      Write the augmented matrix for the system of equations.<\/p>\n

      [latex]\\left[\\begin{array}{rrr}\\hfill 5& \\hfill 3& \\hfill 9\\\\ \\hfill -2& \\hfill 3& \\hfill -1\\\\ \\hfill -1& \\hfill -4& \\hfill 5\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 5\\\\ \\hfill -2\\\\ \\hfill -1\\end{array}\\right][\/latex]<\/div>\n

      On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [latex]\\left[A\\right][\/latex].<\/p>\n

      [latex]\\left[A\\right]=\\left[\\begin{array}{rrrrrrr}\\hfill 5& \\hfill & \\hfill 3& \\hfill & \\hfill 9& \\hfill & \\hfill -1\\\\ \\hfill -2& \\hfill & \\hfill 3& \\hfill & \\hfill -1& \\hfill & \\hfill -2\\\\ \\hfill -1& \\hfill & \\hfill -4& \\hfill & \\hfill 5& \\hfill & \\hfill 1\\end{array}\\right][\/latex]<\/div>\n

      Use the ref(<\/strong> function in the calculator, calling up the matrix variable [latex]\\left[A\\right][\/latex].<\/p>\n

      [latex]\\text{ref}\\left(\\left[A\\right]\\right)[\/latex]<\/div>\n

      Evaluate.<\/p>\n

      [latex]\\begin{array}{l}\\hfill \\\\ \\left[\\begin{array}{rrrr}\\hfill 1& \\hfill \\frac{3}{5}& \\hfill \\frac{9}{5}& \\hfill \\frac{1}{5}\\\\ \\hfill 0& \\hfill 1& \\hfill \\frac{13}{21}& \\hfill -\\frac{4}{7}\\\\ \\hfill 0& \\hfill 0& \\hfill 1& \\hfill -\\frac{24}{187}\\end{array}\\right]\\to \\begin{array}{l}x+\\frac{3}{5}y+\\frac{9}{5}z=-\\frac{1}{5}\\hfill \\\\ \\text{ }y+\\frac{13}{21}z=-\\frac{4}{7}\\hfill \\\\ \\text{ }z=-\\frac{24}{187}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n

      Using back-substitution, the solution is [latex]\\left(\\frac{61}{187},-\\frac{92}{187},-\\frac{24}{187}\\right)[\/latex].<\/p>\n<\/div>\n

      \n

      Example 9: Applying 2 \u00d7 2 Matrices to Finance<\/h3>\n

      Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      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

      [latex]\\begin{array}{l}\\text{ }x+y=12,000\\hfill \\\\ 0.105x+0.12y=1,335\\hfill \\end{array}[\/latex]<\/div>\n

      As a matrix, we have<\/p>\n

      [latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0.105& \\hfill 0.12\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 1,335\\end{array}\\right][\/latex]<\/div>\n

      Multiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.<\/p>\n

      [latex]\\left[\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill 0.015\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 75\\end{array}\\right][\/latex]<\/div>\n

      Then,<\/p>\n

      [latex]\\begin{array}{l}0.015y=75\\hfill \\\\ \\text{ }y=5,000\\hfill \\end{array}[\/latex]<\/div>\n

      So [latex]12,000 - 5,000=7,000[\/latex].<\/p>\n

      Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.<\/p>\n<\/div>\n

      \n

      Example 10: Applying 3 \u00d7 3 Matrices to Finance<\/h3>\n

      Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      We have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at 5% interest, let [latex]y[\/latex] be the amount invested at 8% interest, and let [latex]z[\/latex] be the amount invested at 9% interest. Thus,<\/p>\n

      [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]<\/div>\n

      As a matrix, we have<\/p>\n

      [latex]\\left[\\begin{array}{rrr}\\hfill 1& \\hfill 1& \\hfill 1\\\\ \\hfill 0.05& \\hfill 0.08& \\hfill 0.09\\\\ \\hfill 2& \\hfill 0& \\hfill -1\\end{array}\\text{ }|\\text{ }\\begin{array}{r}\\hfill 10,000\\\\ \\hfill 770\\\\ \\hfill 0\\end{array}\\right][\/latex]<\/div>\n

      Now, we perform Gaussian elimination to achieve row-echelon form.<\/p>\n

      [latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ -0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0.03& \\hfill & \\hfill 0.04& \\hfill \\\\ \\hfill 2& \\hfill & \\hfill 0& \\hfill & \\hfill -1& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 270\\\\ \\hfill & \\hfill 0\\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ -2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0.03& \\hfill & \\hfill 0.04& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -2& \\hfill & \\hfill -3& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 270\\\\ \\hfill & \\hfill -20,000\\end{array}\\right]\\hfill \\\\ \\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\begin{array}{rrrrrr}\\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill \\frac{4}{3}& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -2& \\hfill & \\hfill -3& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 9,000\\\\ \\hfill & \\hfill -20,000\\end{array}\\right]\\hfill \\\\ 2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill \\frac{4}{3}& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill -\\frac{1}{3}& \\hfill \\end{array}|\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 9,000\\\\ \\hfill & \\hfill -2,000\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n

      The third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].<\/p>\n

      The second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex]. Substituting [latex]z=6,000[\/latex], we get<\/p>\n

      [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]<\/div>\n

      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<\/p>\n

      [latex]\\begin{array}{l}x+1,000+6,000=10,000\\hfill \\\\ \\text{ }x=3,000\\text{ }\\hfill \\end{array}[\/latex]<\/div>\n

      The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.<\/p>\n<\/div>\n

      \n

      Try It 6<\/h3>\n

      A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate.<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

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