Chapter 16.9, Problem 1OE

To determine

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

  $\begin{array}{c}x+y+z=2\\ x+2y+2z=-4\\ 2x+2y-z=1\end{array}$

Answer to Problem 1OE

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

Explanation of Solution

Given information:

The system of equations are

  $\begin{array}{c}x+y+z=2\\ x+2y+2z=-4\\ 2x+2y-z=1\end{array}$

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 is the coefficients of the variables in column form,

  $\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$

Calculation:

Consider the system of equations are

  $\begin{array}{c}x+y+z=2\\ x+2y+2z=-4\\ 2x+2y-z=1\end{array}$

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 is the coefficients of the variables in column form,

  $\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$

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 coefficients of z in all three equations.

  $D=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 2\\ 2& 2& -1\end{array}\right|$

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