Chapter 7, Problem 12PS

To determine

Find the values of a,b,c for the given linear system

Answer to Problem 12PS

The values of the constants are $a=12,b=–4,c=10$

Explanation of Solution

Given:

Linear equations:

  $\begin{array}{l}x+2y–3z=a\hfill \\ –x–y+z=b\hfill \\ 2x+3y–2z=c\hfill \end{array}$

Since $\left(–1,2,–3\right)$ is a solution of the equations, this point will satisfy all the three equations, then

  $\begin{array}{l}a=x+2y–3z\hfill \\ =\left(–1\right)+2\times 2–3\left(–3\right)\hfill \\ =–1+4+9\hfill \end{array} =12$

  $\begin{array}{l}b=–x–y+z\hfill \\ =–\left(–1\right)–2–3\hfill \\ =1–2–3\hfill \\ =–4\hfill \end{array}$

  $\begin{array}{l}c=2x+3y–2z\hfill \\ =2\left(–1\right)+3\times 2–2\left(–3\right)\hfill \\ =–2+6+6\hfill \\ =10\hfill \end{array}$

Thus, the values of the constants are $a=12,b=–4,c=10$