Q. 395 - Intermediate Algebra

Question

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

$\left\{\begin{array}{l}3x+4y-3z=-2\\ 2x+3y-z=-12\\ x+y-2z=6\end{array}\right.$

Step-by-Step Solution

Answer

The given system of equations has no solution.

Step 1. Given information

The given system of equations is:  

$\left\{\begin{array}{l}3x+4y-3z=-2\\ 2x+3y-z=-12\\ x+y-2z=6\end{array}\right.$

Step 2. Evaluating the determinant D

In determinant $D$ all the coefficients are taken.

So,

$$\begin{gathered} D=\left|\begin{array}{llr}3& 4& -3\\ 2& 3& -1\\ 1& 1& -2\end{array}\right| \\ D=3(-6+1)-4(-4+1)-3(2-3) \\ D=-15+12+3 \\ D=0 \end{gathered}$$

Step 3. Evaluating the determinant D x

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

So,

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-2& 4& -3\\ -12& 3& -1\\ 6& 1& -2\end{array}\right| \\ {D}_x=3(24+6)+2(-4+1)-3(12+12) \\ {D}_x=90-6-72 \\ {D}_x=12 \end{gathered}$$

Step 4. Evaluating the determinant D y

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

So,

$$\begin{gathered} D_y=\left|\begin{array}{ccc}3& -2& -3\\ 2& -12& -1\\ 1& 6& -2\end{array}\right| \\ {D}_y=3(24+6)+2(-4+1)-3(12+12) \\ {D}_y=90-6-72 \\ {D}_y=12 \end{gathered}$$ 

Step 5. Evaluating the determinant D z

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

So,

$$\begin{gathered} D_z=\left|\begin{array}{ccc}3& 4& -2\\ 2& 3& -12\\ 1& 1& 6\end{array}\right| \\ {D}_z=3(18+12)-4(12+12)-2(2-3) \\ {D}_z=90-96+2 \\ {D}_z=-4 \end{gathered}$$

Step 6. finding solution

We can see that $D=0$ and $D_x,D_y$ and $D_z$ is not equal to zero. So the system is inconsistent and has no solution.