Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 12 - Matrices - 12-4 Inverse Matrices and Systems - Practice and Problem-Solving Exercises - Page 797: 14

Answer

Coefficient Matrix $=\begin{bmatrix} 1 & 3 &-1 \\ 1 & 0 & 2 \\ 0 &2 &-1\end{bmatrix} $; Variable Matrix $=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$; and Constant Matrix $=\begin{bmatrix} 2 \\ 8 \\ 1 \end{bmatrix}$ Our matrix equation is: $\begin{bmatrix} 1 & 3 &-1 \\ 1 & 0 & 2 \\ 0 &2 &-1\end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} =\begin{bmatrix} 2 \\ 8 \\1 \end{bmatrix}$

Work Step by Step

Write the given system of equations into the matrix form in order to label the parts of the matrix equation. Re-arrange the given system of equations as follows: $$x+3y-z=2 \\ x+(0) y+2z=8 \\ 0 x+2y-z=1$$ Our matrix equation is: $\begin{bmatrix} 1 & 3 &-1 \\ 1 & 0 & 2 \\ 0 &2 &-1\end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} =\begin{bmatrix} 2 \\ 8 \\1 \end{bmatrix}$ Our required results are: Coefficient Matrix $=\begin{bmatrix} 1 & 3 &-1 \\ 1 & 0 & 2 \\ 0 &2 &-1\end{bmatrix} $; Variable Matrix $=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$; and Constant Matrix $=\begin{bmatrix} 2 \\ 8 \\ 1 \end{bmatrix}$
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