We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
-3
50
4
avatar+64 

 

Solve the system of equations
\(\begin{align*} x+y+2z & = 1, \\ -2x +4y + 2z & = 1, \\ y +z &= 1. \end{align*}\)

 Jun 22, 2019
 #1
avatar+7683 
+2

\(\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ -2&4&2&|1 \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&6&6&|3\\ \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&1&1&|\dfrac{1}2\\ \end{pmatrix}\\\)

No solution.

 Jun 22, 2019
 #2
avatar+64 
-3

oh god.

CuteDramione  Jun 22, 2019
 #3
avatar+4296 
+1

The best thing to do here is to check if the determinant is equal to or not equal to zero in the three systems of equations.

 

Thus, we have the matrix \(\begin{bmatrix} 1 & 1 & 2\\ -2 & 4 & 2\\ 0 & 1 & 1 \end{bmatrix}\), and it's determinant, which is \(D\), is thus \(1\begin{vmatrix} 4 & 2\\ 1 & 1 \end{vmatrix} - 1\begin{vmatrix} -2 & 2\\ 0 & 1 \end{vmatrix} + 2\begin{vmatrix} -2 & 4 \\ 0 & 1 \end{vmatrix}\), and this is equal to \(0\), so the three systems of equations have no solutions.

 

-tertre

 Jun 22, 2019
 #4
avatar+64 
-4

:O ok.

CuteDramione  Jun 22, 2019

6 Online Users

avatar