Chapter 11.3, Problem 40AYU

$\{\begin{array}{l}x+4y-3z=0\\ 3x-y+3z=0\\ x+y+6z=0\end{array}$

To determine

To find: The solution to the system of equations:

$x\text{ + 4}y-\text{ 3}z=\text{ 0}$

$3x-y\text{ + 3}z=\text{ 0}$

$x\text{ + }y\text{ + 6}z=\text{ 0}$

Answer to Problem 40AYU

$x=\text{ 0};y=\text{ 0};z=\text{ 0}$

Explanation of Solution

Given:

$x\text{ + 4}y-\text{ 3}z=\text{ 0}$

$3x-y\text{ + 3}z=\text{ 0}$

$x\text{ + }y\text{ + 6}z=\text{ 0}$

Formula used:

Cramer’s rule states that:

The solution of a system of three linear equations $a_{1}x+b_{1}y+c_{1}z=l;a_{2}x+b_{2}y+c_{2}z=m;a_{3}x+b_{3}y+c_{3}z=n$ is given by,

$x=\frac{D_x}{D};y=\frac{D_y}{D};z=\frac{D_z}{D}$

Where we have,

$D=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|;D_x=\left|\begin{array}{ccc}l& b_1& c_1\\ m& b_2& c_2\\ n& b_3& c_3\end{array}\right|;D_y=\left|\begin{array}{ccc}{a}_1& l& c_1\\ a_2& m& c_2\\ a_3& n& c_3\end{array}\right|;D_z=\left|\begin{array}{ccc}{a}_1& b_1& l\\ a_2& b_2& m\\ a_3& b_3& n\end{array}\right|$

provided $D\ne 0$.

When $D=0$, the equations are inconsistent or dependent.

Calculation:

$x\text{ + 4}y-\text{ 3}z=\text{ 0}$

$3x-y\text{ + 3}z=\text{ 0}$

$x\text{ + }y\text{ + 6}z=\text{ 0}$

Where we have:

$D=\left|\begin{array}{ccc}1& 4& -3\\ 3& -1& 3\\ 1& 1& 6\end{array}\right|=1\left(-6-3\right)-4\left(18-3\right)-3\left(3+1\right)=1\left(-9\right)-4\left(15\right)-3\left(4\right)=-9-60-12=-81$

$D_x=\left|\begin{array}{ccc}0& 4& -3\\ 0& -1& 3\\ 0& 1& 6\end{array}\right|=0$

$D_y=\left|\begin{array}{ccc}1& 0& -3\\ 3& 0& 3\\ 1& 0& 6\end{array}\right|=0$

$D_z=\left|\begin{array}{ccc}1& 4& 0\\ 3& -1& 0\\ 1& 1& 0\end{array}\right|=0$

So using Cramer’s rule we get,

$x=\frac{D_x}{D}=\frac{0}{-81}=0$

$y=\frac{D_y}{D}=\frac{0}{-81}=0$

$z=\frac{D_z}{D}=\frac{0}{-81}=0$

Thus the solutions are $x=\text{ 0};y=\text{ 0};z=\text{ 0}$.