Question 8.9.5: Use Cramer’s rule to solve the system {x-2z=3 -y+3z=1 2x+5z=......
Use Cramer’s rule to solve the system
\left\{\begin{aligned}x-2 z & =3 \\-y+3 z & =1 \\2 x+5 z & =0\end{aligned}\right.
We shall merely list the various determinants. You should check the results.
\begin{array}{lll}|D| & \left|\begin{array}{rrr}1 & 0 & -2 \\0 & -1 & 3 \\2 & 0 & 5\end{array}\right|=-9, & \left|D_x\right|=\left|\begin{array}{rrr}3 & 0 & -2 \\1 & -1 & 3 \\0 & 0 & 5\end{array}\right|=-15 \\\left|D_y\right| & =\left|\begin{array}{rrr}1 & 3 & -2 \\0 & 1 & 3 \\2 & 0 & 5\end{array}\right|=27, & \left|D_z\right|=\left|\begin{array}{rrr}1 & 0 & 3 \\0 & -1 & 1 \\2 & 0 & 0\end{array}\right|=6\end{array}
By Cramer’s rule, the solution is
x=\frac{\left|D_x\right|}{|D|}=\frac{-15}{-9}=\frac{5}{3}, \quad y=\frac{\left|D_y\right|}{|D|}=\frac{27}{-9}=-3, \quad z=\frac{\left|D_z\right|}{|D|}=\frac{6}{-9}=-\frac{2}{3} .Related Answered Questions
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