Solve the following system of equations by using Gaussian Elimination. That is, solve the system by (a) Triangularizing the System and (b) Solving by Back-Substitution
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Let's consider the following system of equations as an example: 1. \( 2x + 3y + z = 1 \) 2. \( 4x + y + 2z = 2 \) 3. \( 3x + 2y + 3z = 3 \) We can represent this system in augmented matrix form: \[ \begin{bmatrix} 2 & 3 & 1 & | & 1 \\ 4 & 1 & 2 & | & 2 \\ 3 & 2 Show more…
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3. Solve the following systems of equations using Gaussian elimination. (a) 2x1 - 3x2 + 2x3 = 0 x1 - x2 + x3 = 7 -x1 + 5x2 + 4x3 = 4 (b) -x1 - x2 + x3 = 2 2x1 + 2x2 - 4x3 = -4 x1 - 2x2 + 3x3 = 5 (c) -x1 - 3x2 + 2x3 = -2 2x1 + x2 + 3x3 = 9/2 5x1 + 4x2 + 6x3 = 12 (d) 2x1 + 4x2 - 2x3 = 4 x1 - 2x2 - 4x3 = -1 -2x1 - x2 - 3x3 = -4 (e) x1 + x2 + 4x3 = 4 2x1 + x2 + 3x3 = 5 5x1 + 2x2 + 5x3 = 11 (f) 2x1 - 3x2 - x3 = 2 3x1 - 5x2 - 2x3 = -1 9x1 + 6x2 + 4x3 = 1
Philip L.
For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions. $$ \begin{array}{ll}{\text { (a) } x_{1}-2 x_{2}=3} & {\text { (b) } 2 x_{1}-3 x_{2}=5} \\ {2 x_{1}-x_{2}=9} & {-4 x_{1}+6 x_{2}=8}\end{array} $$ $$ \begin{array}{rlrl} \text { (c) } {x_{1}+x_{2}} & {=0} & {\text { (d) } 3 x_{1}+2 x_{2}-x_{3}} & {=4} \\ {2 x_{1}+3 x_{2}} & {=0} & {x_{1}-2 x_{2}+2 x_{3}} & {=1} \\ {3 x_{1}-2 x_{2}} & {=0} & {11 x_{1}+2 x_{2}+x_{3}} & {=14}\end{array} $$ $$ \begin{aligned} \text { (e) } 2 x_{1}+3 x_{2}+x_{3} &=1 \\ x_{1}+x_{2}+x_{3} &=3 \\ 3 x_{1}+4 x_{2}+2 x_{3} &=4 \end{aligned} $$ $$ \begin{aligned} \text { (f) } x_{1}-x_{2}+2 x_{3} &=4 \\ 2 x_{1}+3 x_{2}-x_{3} &=1 \\ 7 x_{1}+3 x_{2}+4 x_{3} &=7 \end{aligned} $$ $$ \begin{array}{r} \text { (g) }{x_{1}+x_{2}+x_{3}+x_{4}=0} \\ {2 x_{1}+3 x_{2}-x_{3}-x_{4}=2} \\ {3 x_{1}+2 x_{2}+x_{3}+x_{4}=5} \\ {3 x_{1}+6 x_{2}-x_{3}-x_{4}=4}\end{array} $$ $$ \begin{aligned} \text { (h) } x_{1}-2 x_{2} &=3 \\ 2 x_{1}+x_{2} &=1 \\-5 x_{1}+8 x_{2} &=4 \end{aligned} $$ $$ \begin{aligned} \text { (i) }-x_{1}+2 x_{2}-x_{3}=& 2 \\-2 x_{1}+2 x_{2}+x_{3}=& 4 \\ 3 x_{1}+2 x_{2}+2 x_{3}=& 5 \\-3 x_{1}+8 x_{2}+5 x_{3}=& 17 \end{aligned} $$ $$ \begin{aligned} \text { (j) } & x_{1}+2 x_{2}-3 x_{3}+x_{4}=1 \\ &-x_{1}-x_{2}+4 x_{3}-x_{4}=6 \\ &-2 x_{1}-4 x_{2}+7 x_{3}-x_{4}=1 \end{aligned} $$ $$ \begin{array}{l}{\text { (k) } x_{1}+3 x_{2}+x_{3}+x_{4}=3} \\ {\quad 2 x_{1}-2 x_{2}+x_{3}+2 x_{4}=8} \\ {\quad x_{1}-5 x_{2} \quad+\quad x_{4}=5}\end{array} $$ $$ \begin{aligned} \text { (l) } x_{1}-3 x_{2}+x_{3} &=1 \\ 2 x_{1}+x_{2}-x_{3} &=2 \\ x_{1}+4 x_{2}-2 x_{3} &=1 \\ 5 x_{1}-8 x_{2}+2 x_{3} &=5 \end{aligned} $$
Matrices and Systems of Equations
Row Echelon Form
Write the following system of equations as an augmented matrix: 3x + 3y + 6z = 12 x - 3y + 5z = 5 2x + 10y - 3z = 0 (a) Reduce the augmented matrix to upper triangular form, then solve by back substitution. (b) Solve by Gaussian elimination. (c) Solve by Gauss-Jordan method.
Adi S.
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