Chapter 16.9, Problem 10OE

To determine

To calculate: The entries of the determinant $D_z$ for the system of equations,

  $\begin{array}{c}-x+3y+2z=4\\ 2x-y+z=2\\ 4x+2y-2z=0\end{array}$

Answer to Problem 10OE

The column one of determinant is $\left[\begin{array}{c}-1\\ 2\\ 4\end{array}\right]$ , column two is $\left[\begin{array}{c}3\\ -1\\ 2\end{array}\right]$ and column three is $\left[\begin{array}{c}4\\ 2\\ 0\end{array}\right]$ .

Explanation of Solution

Given information:

The system of equations are

  $\begin{array}{c}-x+3y+2z=4\\ 2x-y+z=2\\ 4x+2y-2z=0\end{array}$

Formula used:

For system of equations

  $\begin{array}{c}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$

The determinant $D_z$ coefficients of the variables x and y along with constant term in column form,

  $\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$

Calculation:

Consider the system of equations are

  $\begin{array}{c}-x+3y+2z=4\\ 2x-y+z=2\\ 4x+2y-2z=0\end{array}$

Recall that for system of equations

  $\begin{array}{c}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$

The determinant $D_z$ coefficients of the variables x and y along with constant term in column form,

  $\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$

Column one of determinant has coefficients of x in all three equations.

Column two of determinant has coefficients of y in all three equations.

Column three of determinant has constant term of all three equations.

  $D_z=\left|\begin{array}{ccc}-1& 3& 4\\ 2& -1& 2\\ 4& 2& 0\end{array}\right|$

Thus, the column one of determinant is $\left[\begin{array}{c}-1\\ 2\\ 4\end{array}\right]$ , column two is $\left[\begin{array}{c}3\\ -1\\ 2\end{array}\right]$ and column three is $\left[\begin{array}{c}4\\ 2\\ 0\end{array}\right]$ .