Q. 394 - Intermediate Algebra

Question

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

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

Step-by-Step Solution

Answer

This system of equations has infinitely many solutions.

Step 1. Given information

The given system of equations is: 

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

Step 2. Evaluating the determinant D

In determinant $D$ all the coefficients are taken.

So,

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 1& -3\\ 0& 1& -1\\ -1& 2& 0\end{array}\right| \\ D=1(0+2)-1(0-1)-3(0+1) \\ D=2+1-3 \\ D=0 \end{gathered}$$

Step 3. Evaluating the determinant D x

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

So,

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-1& 1& -3\\ 0& 1& -1\\ 1& 2& 0\end{array}\right| \\ {D}_x=-1(0+2)-1(0+1)-3(0-1) \\ {D}_x=-2-1+3 \\ {D}_x=0 \end{gathered}$$

Step 4. Evaluating the determinant D y

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

So,

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& -1& -3\\ 0& 0& -1\\ -1& 1& 0\end{array}\right| \\ {D}_y=1(0+1)+1(0-1)-3(0+0) \\ {D}_y=1-1 \\ {D}_y=0 \end{gathered}$$

Step 5. Evaluating the determinant D z

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

So,

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& 1& -1\\ 0& 1& 0\\ -1& 2& 1\end{array}\right| \\ {D}_z=1(1-0)-(0-0)-1(0+1) \\ {D}_z=0 \end{gathered}$$

We can see that and $D_z$ are zero. So the system of equations is consistent and has infinitely many solutions.