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: 15

Answer

Coefficient Matrix $=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 1 \\ 0 & 1 & 3 \end{bmatrix} $; Variable Matrix $=\begin{bmatrix} r \\ s \\ t \end{bmatrix}$; and Constant Matrix $=\begin{bmatrix} 150 \\ 425 \\ 0 \end{bmatrix}$ Our matrix equation is: $\begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 1 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} r \\ s \\ t \end{bmatrix} =\begin{bmatrix} 150 \\ 425 \\ 0 \end{bmatrix}$

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: $$r-s+t=150\\ 2r+(0) s+t=425 \\ (0)r+s+3t=0$$ Our matrix equation is: $\begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 1 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} r \\ s \\ t \end{bmatrix} =\begin{bmatrix} 150 \\ 425 \\ 0 \end{bmatrix}$ Our required results are: Coefficient Matrix $=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 1 \\ 0 & 1 & 3 \end{bmatrix} $; Variable Matrix $=\begin{bmatrix} r \\ s \\ t \end{bmatrix}$; and Constant Matrix $=\begin{bmatrix} 150 \\ 425 \\ 0 \end{bmatrix}$
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