## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 917: 3

#### Answer

a) The order of the matrix is $3\times 4$. b) The identification of the element ${{a}_{32}}$ is not possible because of ${{a}_{32}}=\frac{1}{2}\text{ and }{{a}_{23}}=\frac{1}{2}$.

#### Work Step by Step

(a) Since this matrix has three rows and four columns, therefore the order of the matrix is $3\times 4$. The order of the matrix $A$ is $3\times 4$. (b) The notation ${{a}_{23}}$ stands for the element present in the second row and third column of the given matrix. Therefore, ${{a}_{23}}=-6$ Also, the notation ${{a}_{32}}$ stands for the element present in the third row and second column of the given matrix. Therefore, ${{a}_{32}}=\frac{1}{2}$ Hence the identification of the element ${{a}_{32}}$ is not possible because of ${{a}_{32}}=\frac{1}{2}\text{ and }{{a}_{23}}=\frac{1}{2}$.

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