Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Test - Page 951: 4

Answer

The matrix is, $\left[ \begin{array}{*{35}{r}} 5 & -2 \\ 1 & -1 \\ 4 & -1 \\ \end{array} \right]$

Work Step by Step

Matrix multiplication: consider two matrices ${{X}_{2\times 2}},{{Y}_{2\times 2}}$. A matrix multiplication is possible only if the rows of the first matrix are equal to the columns of the second matrix and the resulting matrix formed will have the order as: rows of the second matrix and columns of the first matrix. $X=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]$ and $Y=\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right]$ Now, $\begin{align} & X\times Y=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]\times \left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{x}_{1}}{{y}_{1}}+{{x}_{2}}{{y}_{3}} & {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{4}} \\ {{x}_{3}}{{y}_{1}}+{{x}_{4}}{{y}_{3}} & {{x}_{3}}{{y}_{2}}+{{x}_{4}}{{y}_{4}} \\ \end{matrix} \right] \end{align}$ Using the above method, we will solve the given expression as below: $\begin{align} & AB=\left[ \begin{array}{*{35}{r}} 3 & 1 \\ 1 & 0 \\ 2 & 1 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 1 \\ \end{array} \right] \\ & =\left[ \begin{matrix} 3+2 & -3+1 \\ 1+0 & -1+0 \\ 2+2 & -2+1 \\ \end{matrix} \right] \\ & =\left[ \begin{array}{*{35}{r}} 5 & -2 \\ 1 & -1 \\ 4 & -1 \\ \end{array} \right] \end{align}$ Therefore, $AB=\left[ \begin{array}{*{35}{r}} 5 & -2 \\ 1 & -1 \\ 4 & -1 \\ \end{array} \right]$.
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