## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 946: 65

#### Answer

The solution of the matrix $A=\left[ \begin{matrix} 8 & 2 & 6 & -1 & 0 \\ 2 & 0 & -3 & 4 & 7 \\ 2 & 1 & -3 & 6 & -5 \\ -1 & 2 & 1 & 5 & -1 \\ 4 & 5 & -2 & 3 & -8 \\ \end{matrix} \right]$ is $13200$. $\det \left( A \right)\text{ }13200$

#### Work Step by Step

Consider the given matrix: $A=\left[ \begin{matrix} 8 & 2 & 6 & -1 & 0 \\ 2 & 0 & -3 & 4 & 7 \\ 2 & 1 & -3 & 6 & -5 \\ -1 & 2 & 1 & 5 & -1 \\ 4 & 5 & -2 & 3 & -8 \\ \end{matrix} \right]$ Calculate the determinant of the matrix using the graphing utility and following the steps given below: Enter the entities of the matrices and enter the “MATH” mode. Select determinant mode for calculating the determinant. Press 2nd and then press the matrix button of the graphical calculator. Choose the matrix “A” and press enter. Press the enter key once again to display the value of the determinant on the screen: $\det \left( A \right)\text{ }13200$ Therefore, the determinant of the matrix is shown as 13,200.

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