Chapter 11, Problem 15CT

To determine

The solution of the system of linear equations using matrices,

$\{\begin{array}{l}x+2y+4z=-3\\ 2x+7y+15z=-12\\ 4x+7y+13z=-10\end{array}$.

Answer to Problem 15CT

Solution:

The solution of system of equation $\{\begin{array}{l}x+2y+4z=-3\\ 2x+7y+15z=-12\\ 4x+7y+13z=-10\end{array}$ is $x=1,y=-2,z=0$ or using an ordered triplet $\left(1,-2,0\right)$.

Explanation of Solution

Given information:

The system of linear equations, $\{\begin{array}{l}x+2y+4z=-3\\ 2x+7y+15z=-12\\ 4x+7y+13z=-10\end{array}$.

Explanation:

The augmented matrix of the system is,

$\left[\begin{array}{cccc}1& 2& 4& -3\\ 2& 7& 15& -12\\ 4& 7& 13& -10\end{array}\right]$.

Perform the row operation $R_1=-2r_1+r_2,R_3=-4r_1+r_3$,

$\Rightarrow \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 3& 7& -6\\ 0& -1& -3& 2\end{array}\right]$

Perform the row operation $R_2=-r_3,R_3=r_2$,

$\Rightarrow \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& 3& -2\\ 0& 3& 7& -6\end{array}\right]$

Perform the row operation $R_3=-3r_2+r_3$,

$\Rightarrow \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& 3& -2\\ 0& 0& -2& 0\end{array}\right]$

Perform the row operation $R_3=-\frac{1}{2}{r}_3$,

$\Rightarrow \left[\begin{array}{cccc}1& 2& 4& -3\\ 0& 1& 3& -2\\ 0& 0& -2& 0\end{array}\right]$

The matrix is in row echelon form the last row represent equation $-2z=0$ $\Rightarrow z=0$

Second row represent equation $y+3z=-2$

Third row represent an equation, $x+2y+4z=-3$

Substitute $z=0$ in equation $y+3z=-2$

$\Rightarrow y+3\left(0\right)=-2$

$\Rightarrow y=-2$

Substitute $z=0$ and $y=-2$ in $x+2y+4z=-3$

$\Rightarrow x+2\left(-2\right)+4\left(0\right)=-3$

$\Rightarrow x-4=-3$

$\Rightarrow x=1$

Hence, the solution of system of equation $\{\begin{array}{l}x+2y+4z=-3\\ 2x+7y+15z=-12\\ 4x+7y+13z=-10\end{array}$ is $x=1,y=-2,z=0$ or using an ordered triplet $\left(1,-2,0\right)$.