Question

Find all solutions of this system .
4x + 2y + 3z = 0
3x - y + 2z = 0
x + 2y - z =0

Compare the value of the determinant of the coefficient matrix to zero using the nature of the solution

Answer 1

$1.\\\begin{cases}4x + 2y + 3z = 0\\ 3x - y + 2z = 0\\ x + 2y - z =0 \end{cases}\\\begin{cases}4x + 2y + 3z = 0\\ 3x - y + 2z = 0\\ x = z -2y \end{cases}\\\begin{cases}4x + 2y + 3z = 0\\ 3x - y + 2z = 0\\ x = z -2y \end{cases}\\\begin{cases} 2y + 3z+4(z -2y) = 0\\ - y + 2z+3(z -2y) = 0\\ x = z -2y \end{cases}\\\begin{cases} 7z -6y = 0\\ 5z-7y = 0\\ x = z -2y \end{cases}\\\begin{cases} y = \frac{7z}{6}\\ 5z-7y = 0\\ x = z -2y \end{cases}\\\begin{cases} y = \frac{7z}{6}\\ \frac{-19z}{6} = 0\\ x = z -2y \end{cases}\\\begin{cases} y = \frac{7z}{6}\\ z = 0\\ x = z -2y \end{cases}\\\begin{cases} y = 0\\ z = 0\\ x = 0 \end{cases}\\2.\\\begin{vmatrix} 4 & 2 &3\\ 3 & -1&2\\1&2&-1 \end{vmatrix}=\\=4\cdot(-1)\cdot(-1)+2\cdot2\cdot1+3\cdot2\cdot3-\\-3\cdot(-1)\cdot1-3\cdot2\cdot(-1)-2\cdot2\cdot4=19$