Chapter 11, Problem 20CT

To determine

To calculate: The solution of the system of equations $\{\begin{array}{l}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$ by using Cramer’s rule.

Answer to Problem 20CT

Solution:

The solution of the system of equations $\{\begin{array}{l}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$ by using Cramer’s rule is $x=1,y=-1,z=4$ or using ordered triplet $\left(1,-1,4\right)$.

Explanation of Solution

Given information:

The system of linear equations, $\{\begin{array}{l}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$

Formula used:

Cramer’s rule:

The solution of the system of equations $\{\begin{array}{l}{a}_{11}x+a_{12}y+a_{13}z=c_1\\ a_{21}x+a_{22}y+a_{23}z=c_2\\ a_{31}x+a_{32}y+a_{33}z=c_3\end{array}$ is given by,

$x=\frac{D_x}{D},y=\frac{D_y}{D},z=\frac{D_z}{D}$ provided that $D=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\ne 0$, where $D_x=\left|\begin{array}{ccc}{c}_1& a_{12}& a_{13}\\ c_2& a_{22}& a_{23}\\ c_3& a_{32}& a_{33}\end{array}\right|,D_y=\left|\begin{array}{ccc}{a}_{11}& c_1& a_{13}\\ a_{21}& c_2& a_{23}\\ a_{31}& c_3& a_{33}\end{array}\right|,D_z=\left|\begin{array}{ccc}{a}_{11}& a_{12}& c_1\\ a_{21}& a_{22}& c_2\\ a_{31}& a_{32}& c_3\end{array}\right|$.

Calculation:

Compare given system of equations $\{\begin{array}{l}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$ with $\{\begin{array}{l}{a}_{11}x+a_{12}y+a_{13}z=c_1\\ a_{21}x+a_{22}y+a_{23}z=c_2\\ a_{31}x+a_{32}y+a_{33}z=c_3\end{array}$,

Gives, $a_{11}=4,a_{12}=-3,a_{13}=2,a_{21}=-2,a_{22}=1,a_{23}=-3,a_{31}=5,a_{32}=-5,a_{33}=2$ and $c_1=15,c_2=-15,c_3=18$.

$D=\left|\begin{array}{ccc}4& -3& 2\\ -2& 1& -3\\ 5& -5& 2\end{array}\right|$

$={\left(-1\right)}^{1+1}\cdot \left(4\right)\cdot \left|\begin{array}{cc}1& -3\\ -5& 2\end{array}\right|+{\left(-1\right)}^{1+2}\cdot \left(-3\right)\cdot \left|\begin{array}{cc}-2& -3\\ 5& 2\end{array}\right|+{\left(-1\right)}^{1+3}\cdot \left(2\right)\cdot \left|\begin{array}{cc}-2& 1\\ 5& -5\end{array}\right|$

$=4\cdot \left|\begin{array}{cc}1& 3\\ -5& 2\end{array}\right|+3\cdot \left|\begin{array}{cc}-2& -3\\ 5& 2\end{array}\right|+2\cdot \left|\begin{array}{cc}-2& 1\\ 5& -5\end{array}\right|$

$=4\cdot \left(2-15\right)+3\cdot \left(-4+15\right)+2\cdot \left(10-5\right)$

$=4\cdot \left(-13\right)+3\cdot \left(11\right)+2\cdot \left(5\right)$

$=-52+33+10$

$=-9$

Thus, $D=-9\ne 0$

$D_x=\left|\begin{array}{ccc}{c}_1& a_{12}& a_{13}\\ c_2& a_{22}& a_{23}\\ c_3& a_{32}& a_{33}\end{array}\right|=\left|\begin{array}{ccc}15& -3& 2\\ -15& 1& -3\\ 18& -5& 2\end{array}\right|$

$={\left(-1\right)}^{1+1}\cdot \left(15\right)\cdot \left|\begin{array}{cc}1& -3\\ -5& 2\end{array}\right|+{\left(-1\right)}^{1+2}\cdot \left(-3\right)\cdot \left|\begin{array}{cc}-15& -3\\ 18& 2\end{array}\right|+{\left(-1\right)}^{1+3}\cdot \left(2\right)\cdot \left|\begin{array}{cc}-15& 1\\ 18& -5\end{array}\right|$

$=15\cdot \left|\begin{array}{cc}1& -3\\ -5& 2\end{array}\right|+3\cdot \left|\begin{array}{cc}-15& -3\\ 18& 2\end{array}\right|+2\cdot \left|\begin{array}{cc}-15& 1\\ 18& -5\end{array}\right|$

$=15\cdot \left(2-15\right)+3\cdot \left(-30+54\right)+2\cdot \left(75-18\right)$

$=15\cdot \left(-13\right)+3\cdot \left(24\right)+2\cdot \left(57\right)$

$=-195+72+114$

Thus, $D_x=-9$.

By using Cramer’s rule solution,

$x=\frac{-9}{-9}=1$.

$D_y=\left|\begin{array}{ccc}{a}_{11}& c_1& a_{13}\\ a_{21}& c_2& a_{23}\\ a_{31}& c_3& a_{33}\end{array}\right|=\left|\begin{array}{ccc}4& 15& 2\\ -2& -15& -3\\ 5& 18& 2\end{array}\right|$

$={\left(-1\right)}^{1+1}\cdot \left(4\right)\cdot \left|\begin{array}{cc}-15& -3\\ 18& 2\end{array}\right|+{\left(-1\right)}^{1+2}\cdot \left(15\right)\cdot \left|\begin{array}{cc}-2& -3\\ 5& 2\end{array}\right|+{\left(-1\right)}^{1+3}\cdot \left(2\right)\cdot \left|\begin{array}{cc}-2& -15\\ 5& 18\end{array}\right|$

$=4\cdot \left|\begin{array}{cc}-15& -3\\ 18& 2\end{array}\right|-15\cdot \left|\begin{array}{cc}-2& -3\\ 5& 2\end{array}\right|+2\cdot \left|\begin{array}{cc}-2& -15\\ 5& 18\end{array}\right|$

$=4\cdot \left(-30+54\right)-15\cdot \left(-4+15\right)+2\cdot \left(-36+75\right)$

$=4\cdot \left(24\right)-15\cdot \left(11\right)+2\cdot \left(39\right)$

$=96-165+78$

$=9$

Thus, $D_y=9$.

By using Cramer’s rule solution,

$y=\frac{9}{-9}=-1$.

$D_z=\left|\begin{array}{ccc}{a}_{11}& a_{12}& c_1\\ a_{21}& a_{22}& c_2\\ a_{31}& a_{32}& c_3\end{array}\right|=\left|\begin{array}{ccc}4& -3& 15\\ -2& 1& -15\\ 5& -5& 18\end{array}\right|$

$={\left(-1\right)}^{1+1}\cdot \left(4\right)\cdot \left|\begin{array}{cc}1& -15\\ -5& 18\end{array}\right|+{\left(-1\right)}^{1+2}\cdot \left(-3\right)\cdot \left|\begin{array}{cc}-2& -15\\ 5& 18\end{array}\right|+{\left(-1\right)}^{1+3}\cdot \left(15\right)\cdot \left|\begin{array}{cc}-2& 1\\ 5& -5\end{array}\right|$

$=4\cdot \left|\begin{array}{cc}1& -15\\ -5& 18\end{array}\right|+3\cdot \left|\begin{array}{cc}-2& -15\\ 5& 18\end{array}\right|+15\cdot \left|\begin{array}{cc}-2& 1\\ 5& -5\end{array}\right|$

$=4\cdot \left(18-75\right)+3\cdot \left(-36+75\right)+15\cdot \left(10-5\right)$

$=4\cdot \left(-57\right)+3\cdot \left(39\right)+15\cdot \left(5\right)$

$=-228+117+75$

$=-36$

Thus, $D_z=-36$.

By using Cramer’s rule solution,

$z=\frac{-36}{-9}=4$.

Thus, the solution of the system of equations $\{\begin{array}{l}4x-3y+2z=15\\ -2x+y-3z=-15\\ 5x-5y+2z=18\end{array}$ by using Cramer’s rule is $x=1,y=-1,z=4$ or using ordered triplet $\left(1,-1,4\right)$.