Find the determinant 1 11 11 21 21 12 33 61 3

1 1 1 1 1 2 2 3 1 2 3 6 1 1 1 3 = 1 0 1 1 1 1 2 3 1 1 3 6 1 0 1 3…

Answered: 1 1 1 2 1 2 1 3 1 1 2 3 3 6 1 1

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterA: Appendix

SectionA.3: Inequalities

Problem 11E

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Question

Find the determinant

Expert Solution

Step 1

$|\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 3 \\ 1 & 2 & 3 & 6 \\ 1 & 1 & 1 & 3 \end{matrix}|$

$=$ $|\begin{matrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 2 & 3 \\ 1 & 1 & 3 & 6 \\ 1 & 0 & 1 & 3 \end{matrix}|$ $[{C_{2}}^{'} = C_{2} - C_{1}]$

$=$ $|\begin{matrix} 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 3 \\ 1 & 1 & 2 & 6 \\ 1 & 0 & 0 & 3 \end{matrix}|$ $[{C_{3}}^{'} = C_{3} - C_{1}]$

$=$ $|\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 2 \\ 1 & 1 & 2 & 5 \\ 1 & 0 & 0 & 2 \end{matrix}|$ $[{C_{4}}^{'} = C_{4} - C_{1}]$

Expand along the first row,we get

$|\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 2 \\ 1 & 1 & 2 & 5 \\ 1 & 0 & 0 & 2 \end{matrix}|$ $=$ $1 \cdot {(- 1)}^{1 + 1} |\begin{matrix} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 0 & 0 & 2 \end{matrix}| + 0 \cdot {(- 1)}^{1 + 2} |\begin{matrix} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 1 & 0 & 2 \end{matrix}| + 0 \cdot {(- 1)}^{1 + 3} |\begin{matrix} 1 & 1 & 2 \\ 1 & 1 & 5 \\ 1 & 0 & 2 \end{matrix}| + 0 \cdot {(- 1)}^{1 + 4} |\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 0 \end{matrix}|$

$=$ $|\begin{matrix} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 0 & 0 & 2 \end{matrix}|$

Now expand along the third row of $|\begin{matrix} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 0 & 0 & 2 \end{matrix}|$ , we get

$|\begin{matrix} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 0 & 0 & 2 \end{matrix}|$ $=$ $0 \cdot {(- 1)}^{3 + 1} |\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}| + 0 \cdot {(- 1)}^{3 + 2} |\begin{matrix} 1 & 2 \\ 1 & 5 \end{matrix}| + 2 \cdot {(- 1)}^{3 + 3} |\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}|$

$=$ $2 |\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}|$

The determinant of a $2 \times 2$ of matrix is $|\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}| = 1 \cdot 2 - 1 \cdot 1 = 2 - 1 = 1$

Therefore , $2 |\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}| = 2 \cdot 1 = 2$

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