+73
Solve the system of equations
\(\begin{align*} x+y+2z & = 1, \\ -2x +4y + 2z & = 1, \\ y +z &= 1. \end{align*}\)
+9519 \(\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ -2&4&2&|1 \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&6&6&|3\\ \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&1&1&|\dfrac{1}2\\ \end{pmatrix}\\\)
No solution.
+4609 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
