Q. 392 - Intermediate Algebra

Question

In the following exercise, solve each system of equations using Cramer’s rule

$\left\{\begin{array}{l}4x-3y+z=7\\ 2x-5y-4z=3\\ 3x-2y-2z=-7\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equation is $\left(-3,-5,4\right)$.

Step 1. Given information

The given system of equations is: 

$\left\{\begin{array}{l}4x-3y+z=7\\ 2x-5y-4z=3\\ 3x-2y-2z=-7\end{array}\right.$

Step 2. Evaluating the determinant D

In determinant $D$ all the coefficients are taken.

So,

$$\begin{gathered} D=\left|\begin{array}{rrr}4& -3& 1\\ 2& -5& -4\\ 3& -2& -2\end{array}\right| \\ D=4(10-8)+3(-4+12)+1(-4+15) \\ D=8+24+11 \\ D=43 \end{gathered}$$

Step 3. Evaluating the determinant D x

In determinant $D_x$ we take the constants in place of coefficients of $x$ 

So,

$$\begin{gathered} D_x=\left|\begin{array}{ccc}7& -3& 1\\ 3& -5& -4\\ -7& -2& -2\end{array}\right| \\ {D}_x=7(10-8)+3(-6-28)+1(-6-35) \\ {D}_x=7\left(2\right)+3(-34)+(-41) \\ {D}_x=14-102-41 \\ {D}_x=-129 \end{gathered}$$

Step 4. Evaluating the determinant D y

In the determinant $D_y$, we take the constant in place of coefficients of $y$

So,

$$\begin{gathered} D_y=\left|\begin{array}{ccc}4& 7& 1\\ 2& 3& -4\\ 3& -7& -2\end{array}\right| \\ {D}_y=4(-6-28)-7(-4+12)+1(-14-9) \\ {D}_y=4(-34)-7\left(8\right)-23 \\ {D}_y=-136-56-23 \\ {D}_y=-215 \\ {D}_y= \end{gathered}$$

Step 5. Evaluating the determinant D z

In the determinant  $D_z$, we take the constant in place of coefficients of $z$ 

So,

$D_z=\left|\begin{array}{ccc}4& -3& 7\\ 2& -5& 3\\ 3& -2& -7\end{array}\right|$

$$\begin{gathered} D_z=4(35+6)+3(-14-9)+7(-4+15) \\ {D}_z=164-69+77 \\ {D}_z=172 \end{gathered}$$

Step 6. Finding the value of x , y & z

For $x$

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-129}{43} \\ x=-3 \end{gathered}$$

For $y$

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{-215}{43} \\ y=-5 \end{gathered}$$

For $z$

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{172}{43} \\ z=4 \end{gathered}$$

Step 7. Writing solution in ordered traid

The solution of the system in the ordered triad is $\left(-3,-5,4\right)$

Step 8. Check the solution for the equation 4 x - 3 y + z = 7

Substituting $\left(-3,-5,4\right)$ in the equation, we get:

$$\begin{gathered} 4x-3y+z=7 \\ 4(-3)-3(-5)+4=7 \\ -12+15+4=7 \\ 7=7 \end{gathered}$$

This is true.

Step 9. Check the solution for the equation 2 x   −   5 y   −   4 z   =   3

Substituting  $\left(-3,-5,4\right)$in the equation, we get:

$$\begin{gathered} 2x - 5y - 4z = 3 \\ 2(-3)-5(-5)-4\left(4\right)=3 \\ 3=3 \end{gathered}$$

This is true

Step 10. Check the solution for the equation 3 x   −   2 y   −   2 z   =   − 7

Substituting $3x - 2y - 2z = -7$ in the equation, we get:

$$\begin{gathered} 3x - 2y - 2z = -7 \\ 3(-3)-2(-5)-2\left(4\right)=-7 \\ -7=-7 \end{gathered}$$

This is also true.

So, the ordered pair $\left(-3,-5,4\right)$ is the solution of the given system of the equations.