Q16SE Question: 16. Let J be the \(n \... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 3: Q16SE (page 165)

Question: 16. Let J be the $n \times n$matrix of all 1鈥檚, and consider$A = \left( {a - b} \right)I + bJ$; that is,
$A = \left( {\begin{array}{*{20}{c}}a&b&b& \cdots &b\\b&a&b& \cdots &b\\b&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)$
Confirm that $det\,A = {\left( {a - b} \right)^{n - {\bf{1}}}}\left( {a + \left( {n - {\bf{1}}} \right)b} \right)$ as follows:
Subtract row 2 from row 1, row 3 from row 2, and so on, and explain why this not change the determinant of the matrix.
With the resulting matrix from part (a), add column 1 to column 2, then add this new column 2 to column 3, and so on, and explain why this does not change the determinant.
Find the determinant of the resulting matrix from (b)

Short Answer

Expert verified

Since row replacement operations do not change the determinant, hence, $\det A = \det \left( {\begin{array}{*{20}{c}}{a - b}&{b - a}&0& \cdots &0\\0&{a - b}&{b - a}& \cdots &0\\0&0&{a - b}& \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)$

Since $\det A = \det {A^T}$, so the column replacement operations in $A$ are equivalent to the row replacement operations in ${A^T}$. So, the column replacement operation does not change the determinant. Therefore, $\det A = \det \left( {\begin{array}{*{20}{c}}{a - b}&0&0& \cdots &0\\0&{a - b}&0& \cdots &0\\0&0&{a - b}& \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&{2b}&{3b}& \cdots &{a + \left( {n - 1} \right)b}\end{array}} \right)$

$\det A = {\left( {a - b} \right)^{n - 1}}\left( {a + \left( {n - 1} \right)b} \right)$

Step by step solution

Do the given operations in part (a)

(a)

From the given matrix,

$\det A = \det \left( {\begin{array}{*{20}{c}}a&b&b& \cdots &b\\b&a&b& \cdots &b\\b&b&a& \cdots &b\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)$

At row 1, subtract row 2 from row 1; at row 2, subtract row 3 from row 2, and so on. These are row replacement operations. Hence, these do not change the determinant of the matrix. Thus,

$\det A = \det \left( {\begin{array}{*{20}{c}}{a - b}&{b - a}&0& \cdots &0\\0&{a - b}&{b - a}& \cdots &0\\0&0&{a - b}& \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&b&b& \cdots &a\end{array}} \right)$

Continue from part (a)

(b)

Since $\det A = \det {A^T}$, so the column replacement operations in $A$ are equivalent to the row replacement operations in ${A^T}$. Hence, this does not change the determinant. In the resulting matrix from part (a), at column 2, add column 1 to column 2 to obtain:

$\det A = \det \left( {\begin{array}{*{20}{c}}{a - b}&0&0& \cdots &0\\0&{a - b}&{b - a}& \cdots &0\\0&0&{a - b}& \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&{2b}&b& \cdots &a\end{array}} \right)$

Now, at column 3, add column 2 to column 3 to obtain:

$\det A = \det \left( {\begin{array}{*{20}{c}}{a - b}&0&0& \cdots &0\\0&{a - b}&0& \cdots &0\\0&0&{a - b}& \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\b&{2b}&{3b}& \cdots &a\end{array}} \right)$

Continue this process to obtain:

Find the determinant of the resulting matrix from (b)

Note that the resulting matrix from part (b) is a triangular matrix. Hence, its determinant is given by the product of its diagonal entries. Therefore,

$\begin{array}{c}\det A = \left( {a - b} \right)\left( {a - b} \right)\left( {a - b} \right) \cdots \left( {a - b} \right)\left( {a + \left( {n - 1} \right)b} \right)\\ = {\left( {a - b} \right)^{n - 1}}\left( {a + \left( {n - 1} \right)b} \right)\end{array}$