Question

Solve the system of homogeneous equations:
2x + 3y - z = 0
x - y - 2z = 0
3x + y + 3z = 0

Answer 1

Answer:

$x=0,y=0,z=0$

Step-by-step explanation:

Cramer's Rule:

it says that

$x=\frac{\triangle}{\triangle_{1}}\\\\\\y=\frac{\triangle}{\triangle_{2}}\\\\\\z=\frac{\triangle}{\triangle_{3}}$

where Δ is determinant of matrix A,Δ1,Δ2,Δ3 are the determinant of A when column 1,2,3 are replaced by coefficient matrix respectively.$\triangle=\left|\begin{array}{ccc}2&3&-1\\1&-1&-2\\3&1&3\end{array}\right|=-33$

$\triangle_{1}=\left|\begin{array}{ccc}0&3&-1\\0&-1&-2\\0&1&3\end{array}\right|=0$

$\triangle_{2}=\left|\begin{array}{ccc}2&0&-1\\1&0&-2\\3&0&3\end{array}\right|=0$

$\triangle_{3}=\left|\begin{array}{ccc}2&3&0\\1&-1&0\\3&1&0\end{array}\right|=0$

$x=\frac{0}{-33} ,y=\frac{0}{-33},z=\frac{0}{-33}\\\\x=0,y=0,z=0$

Answer 2

x=0,y=0,z=0

Step-by-step explanation:

Cramer's Rule:

it says that

\begin{gathered}x=\frac{\triangle}{\triangle_{1}}\\\\\\y=\frac{\triangle}{\triangle_{2}}\\\\\\z=\frac{\triangle}{\triangle_{3}}\\\\\end{gathered}x=△1△y=△2△z=△3△

where Δ is determinant of matrix A,Δ1,Δ2,Δ3 are the determinant of A when column 1,2,3 are replaced by coefficient matrix respectively.\begin{gathered}\triangle=\left|\begin{array}{ccc}2&3&-1\\1&-1&-2\\3&1&3\end{array}\right|=-33\\\\\\\end{gathered}△=∣∣∣∣∣∣∣2133−11−1−23∣∣∣∣∣∣∣=−33

\begin{gathered}\triangle_{1}=\left|\begin{array}{ccc}0&3&-1\\0&-1&-2\\0&1&3\end{array}\right|=0\\\\\end{gathered}△1=∣∣∣∣∣∣∣0003−11−1−23∣∣∣∣∣∣∣=0

\begin{gathered}\triangle_{2}=\left|\begin{array}{ccc}2&0&-1\\1&0&-2\\3&0&3\end{array}\right|=0\\\\\\\end{gathered}△2=∣∣∣∣∣∣∣213000−1−23∣∣∣∣∣∣∣=0

\begin{gathered}\triangle_{3}=\left|\begin{array}{ccc}2&3&0\\1&-1&0\\3&1&0\end{array}\right|=0\\\\\\\end{gathered}△3=∣∣∣∣∣∣∣2133−11000∣∣∣∣∣∣∣=0

\begin{gathered}x=\frac{0}{-33} ,y=\frac{0}{-33},z=\frac{0}{-33}\\\\x=0,y=0,z=0\end{gathered}x=−330,y=−330,z=−330x=0,y=0,z=0