Q 4.99 - Intermediate Algebra

Question

Solve using Cramer's rule  $\left\{ \begin{array}{l} 3x+8y+2z=-5\\ 2 x+5y-3z=0\\ x+2y-2z=-1\end{array}\right.$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{linear} \mathrm{equations} \mathrm{is} (x=-9, \\ y=3, \\ z=-1). \end{gathered}$$

Step 1. Given information

The given linear equations are 

$\left\{ \begin{array}{l} 3x+8y+2z=-5\\ 2 x+5y-3z=0\\ x+2y-2z=-1\end{array}\right.$

Step 2. First find the determinant D by using the coefficients of the variables.

$$\begin{gathered} D=\left|\begin{array}{ccc}3& 8& 2\\ 2& 5& -3\\ 1& 2& -2\end{array}\right| \\ D=3\left(\right(5\times -2)-(-3\times 2\left)\right)-8\left(\right(2\times -2)-(-3\times 1\left)\right)-2\left(\right(2\times 2)-(5\times 1\left)\right) \\ D=3(-10+6)-8(-4+3)-2(4-5) \\ D=-12+8-2 \\ D=-6 \end{gathered}$$

Step 3. Evaluate the determinant D x , D y and D z by using the constants to replace the coefficients of x, y and z.

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-5& 8& 2\\ 0& 5& -3\\ -1& 2& -2\end{array}\right| \\ {D}_x=-5\left(\right(5\times -2)-(-3\times 2\left)\right)-8\left(\right(0\times -2)-(-3\times -1\left)\right)+2\left(\right(0\times 2)-(5\times -1\left)\right) \\ {D}_x=-5(-10+6)-8(+3)+2(+5) \\ {D}_x=20+24+10 \\ {D}_x=6 \\ {D}_y=\left|\begin{array}{ccc}3& -5& 2\\ 2& 0& -3\\ 1& -1& -2\end{array}\right| \\ {D}_y=3\left(\right(0\times -2)-(-3\times -1\left)\right)+5\left(\right(2\times -2)-(-3\times 1\left)\right)+2\left(\right(-1\times 2)-(0\times 1\left)\right) \\ {D}_y=3\left(3\right)+5(-4+3)+2(-2) \\ {D}_y=-9-5-4 \\ {D}_y=-18 \\ {D}_z=\left|\begin{array}{ccc}3& 8& -5\\ 2& 5& 0\\ 1& 2& -1\end{array}\right| \\ {D}_z=3\left(\right(5\times -1)-(0\times 2\left)\right)-8\left(\right(2\times -1)-(0\times 1\left)\right)-5\left(\right(2\times 2)-(5\times 1\left)\right) \\ {D}_z=3(-5)-8(-2)-5(4-5) \\ {D}_z=-15+16+5 \\ {D}_z=6 \end{gathered}$$

Step 4. Find x, y and z.

$$\begin{gathered} x=\frac{D_x}{D}, \\ y=\frac{D_y}{D} \mathrm{and} \\ z=\frac{D_z}{D} \\ x=\frac{54}{-6}=-9 \\ y=\frac{-18}{-6}=3 \\ z=\frac{6}{-6}=-1 \end{gathered}$$

The solution of the linear equations is $(-9,3,-1)$.

By putting the values in the equations it satisfies the equations and the solution to these linear equations are $(-9,3,-1)$.

Step 5. Checking the equation

Putting the value of
$$\begin{gathered} x=-9 \\ y=3 \\ z=-1 \end{gathered}$$
in the equation

$$\begin{gathered} 3\mathrm{x}+8\mathrm{y}+2\mathrm{z}=-5 \\ 3(-9)+8\left(3\right)+2(-1)=-5 \\ -27+24-2=-5 \\ -5=-5 \\ True \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{x}=-9, \\ \mathrm{y}=3, \\ \mathrm{z}=-1 \mathrm{in} \mathrm{the} \mathrm{equation} \\ 2x+5y-2z=0 \\ 2(-9)+5\left(3\right)-3(-1)=0 \\ -18+15+3=0 \\ 0=0 \\ \mathrm{True} \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{x}=-9, \\ \mathrm{y}=3, \\ \mathrm{z}=-1 \mathrm{in} \mathrm{the} \mathrm{equation} \\ x+2y-2z=-1 \\ -9+2\left(3\right)-2(-1)=-1 \\ -9+6+2=-1 \\ -1=-1 \end{gathered}$$

True

Hence the solution to the equation is $(-9,3,-1)$