Find all solutions of the following systems using Gaussian elimination. x + 2y + 3z = 4 2x + 5y...

Question:

Find all solutions of the following systems using Gaussian elimination.

x + 2y + 3z = 4

2x + 5y + 7z = 10

2y + 2z = 4

Gauss Elimination

Gauss elimination is a technique to solve a system of linear equations by eliminating the unknowns from the equations in such a manner that the matrix of the system is a upper triangular matrix.

To solve a system that is upper triangular, we just backsubstitute the variables in the upper equations, starting with the lowest equation.

To apply Gauss elimination, we will do row operations on the matrix of the system.

Answer and Explanation:


To solve the system {eq}\displaystyle \begin{align} \begin{cases} x + 2y + 3z = 4\\ 2x + 5y + 7z = 10\\ 2y + 2z = 4\\ \end{cases} \end{align} {/eq}

we will do row operations on the matrix of the system, together with the right hand side, until we obtain zero under the main diagonal of the matrix.

We will use the notation R_i for the i-th row and describe the operations on the rows on top of the equivalent symbol.

{eq}\displaystyle \begin{align} \left[\begin{array}{ccc | c} 1 & 2 &3 & 4\\ 2&5&7 & 10 \\ 0&2&2 & 4 \end{array}\right] \overset{-2\cdot R_1+R_2}{\iff} \left[\begin{array}{ccc | c} 1 & 2 &3 & 4\\ 0 & 1 &1 & 2\\ 0 & 2 &2 & 4 \end{array}\right] \overset{-2\cdot R_2+R_3}{\iff} \left[\begin{array}{ccc | c} 1 & 2 &3 & 4\\ 0 & 1 &1 & 2\\ 0 & 0 &0 & 0 \end{array}\right] \end{align} {/eq}

We obtained an upper triangular matrix, meaning we have zeros below the diagonal terms, next, we will give the solution by writing the systems of equations corresponding to the last matrix.

Note, by doing the above row operations, the solution of the system is not changed.

{eq}\displaystyle \begin{align} &\begin{cases} \begin{array}{cccc} x & +2y &+3z & =4\\ & y &+z &= 2 \\ & &0 &= 0 \end{array} \end{cases} \\\\ \iff &\begin{cases} \begin{array}{cccc} x & +2y &+3z & =4\\ & y &+z &= 2\\ & &0 &= 0 &\iff &z&=\lambda -\text{ free variable} \end{array} \end{cases} \\\\ \iff &\begin{cases} \begin{array}{cccc} x & +2y &+3z & =4\\ & y &+z &= 2 &\iff& y&=&2-\lambda, \text{ used in the first equation:}\\ & &0 &= 0 &\iff &z&=&\lambda -\text{ free variable} \end{array} \end{cases}\\\\ \iff &\begin{cases} \begin{array}{cccc} x & +2y &+3z & =4 & \iff & x&=&4-2y-3z=-\lambda\\ & y &+z &= 2 &\iff & y&=&2-\lambda\\ & &0 &= 0 &\iff &z&=&\lambda -\text{ free variable} \end{array} \end{cases} \end{align} {/eq}

Therefore, {eq}\displaystyle \begin{align} \boxed{\begin{cases} \begin{array}{cccc} x & =&-\lambda \\ y &=& 2-\lambda \\ z&=& \lambda & -\text{ any real number} \end{array} \end{cases}} \end{align} {/eq} Thus, the system of equations have infinitely many solutions.


Learn more about this topic:

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How to Solve Linear Systems Using Gaussian Elimination

from Algebra II Textbook

Chapter 10 / Lesson 6
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