# 1. The linear system x-y+2z =1, x+2y+z =2, 3r+5z = 4 has infinitely many solutions. Give a...

## Question:

1. The linear system {eq}x-y+2z =1{/eq}, {eq}x+2y+z =2{/eq}, {eq}3r+5z = 4{/eq} has infinitely many solutions. Give a parametric representation of the solution set, using {eq}k{/eq} as the parameter and setting {eq}z= k{/eq} after you have reduced the system to reduced echelon form.

2. Show that the following linear system has no solutions by using elimination to reduce to row echelon form and showing that an incorrect identity results {eq}x+2y+z=3{/eq}, {eq}x-y+2z=1{/eq}, {eq}2x+y+3z=2{/eq}.

## Systems of Equations

To solve a system of linear equations, we can apply Gaussian elimination by reducing the number of unknowns from 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 back substitute 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.

1. To solve the system {eq}\displaystyle \begin{align} \begin{cases} x -y + 2z = 1\\ x + 2y + z =2 \\ x + 5z = 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. This is Gaussian elimination technique.

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

{eq}\displaystyle \begin{align} &\begin{cases} \begin{array}{cccc} x & -y &+2z & =1\\ &3 y &-z &= 1 \\ & &0 &= 0 \end{array} \end{cases} \\\\ \iff &\begin{cases} \begin{array}{cccc} x & -y &+2z & =1\\ &3 y &-z &= 1 \\ & &0 &= 0 &\iff &z =\lambda -\text{ free variable, which used in the upper equations, } \end{array} \end{cases} \\\\ \iff &\begin{cases} \begin{array}{cccc} x & -y &+2z & =&1&\\ &3 y &-z &= &1 & \implies &y=\frac{1}{3}+\frac{\lambda}{3 }\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 & -y &+2z & =1& \implies& x&\displaystyle=&\frac{4}{3}-\frac{5\lambda}{3 } \\ &3 y &-z &= 1 &\implies & y&\displaystyle =&\frac{1}{3}+\frac{\lambda}{3 }\\ & &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 & =&\displaystyle\frac{4}{3}-\frac{5\lambda}{3 } \\ y &=& \displaystyle \frac{1}{3}+\frac{\lambda}{3 } \\ z&=& \lambda & -\text{ any real number} \end{array} \end{cases}} \end{align} {/eq}

Thus, the system of equations have infinitely many solutions.

2. To solve the system {eq}\displaystyle \begin{align} \begin{cases} x -y + 2z = 1\\ x + 2y + z =3 \\ 2x + y+3z = 2\\ \end{cases} \end{align} {/eq}

we will do row operations on the matrix of the system, as before.

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

{eq}\displaystyle \begin{align} &\begin{cases} \begin{array}{cccc} x & -y &+2z & =&1&\\ &3 y &-z &= &1& \\ & &0 &=& 1 &\implies \boxed{\text{ inconsistent , so, there is no solution of the system} } \end{array} \end{cases} \end{align} {/eq}