Consider the 10 × 10 matrix Bt =see photo3a: Calculate its determinant det(Bt) as a function of t ∈ R3b: For which values t ∈ R is the matrix B not invertible? [Why?] 3: Consider the 10 x 10 matrix B, =(10 -(9 – t)10(8 - t)10(7 - t)9.10(6 - t)79.10(5 – t)7810(4 -t)7810(3 - t)467810(2 - t)37810(1 - t)347103a: Calculate its determinant det (B.) as a function of t€R.3b: For which values t e R is the matrix B not invertible? [Why?]

Answered: Consider the 10 × 10 matrix Bt =see…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Consider the 10 × 10 matrix Bt =see photo

3a: Calculate its determinant det(Bt) as a function of t ∈ R

3b: For which values t ∈ R is the matrix B not invertible? [Why?]

3: Consider the 10 x 10 matrix B, =
(10 -
(9 – t)
10
(8 - t)
10
(7 - t)
9.
10
(6 - t)
7
9.
10
(5 – t)
7
8
10
(4 -t)
7
8
10
(3 - t)
4
6
7
8
10
(2 - t)
3
7
8
10
(1 - t)
3
4
7
10
3a: Calculate its determinant det (B.) as a function of t€R.
3b: For which values t e R is the matrix B not invertible? [Why?]

Expert Solution

Step 1

For part 3a:

To find the determinant of given matrix as a function of t ∈ R.

Given matrix:

$B_{t} = [\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & (10 - t) \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & (9 - t) & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & (8 - t) & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & (7 - t) & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & (6 - t) & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & (5 - t) & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & (4 - t) & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & (3 - t) & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & (2 - t) & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ (1 - t) & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \end{matrix}]$

Step 2

Note that interchanging two rows or two columns of matrix reverse the sign of it's determinant from positive to negative or from negative to positive.

Here to find the determinant of given matrix first convert it into a upper triangular matrix, then it's determinant will be product of it's diagonal matrix of the new matrix formed.

$|\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & (10 - t) \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & (9 - t) & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & (8 - t) & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & (7 - t) & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & (6 - t) & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & (5 - t) & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & (4 - t) & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & (3 - t) & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & (2 - t) & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ (1 - t) & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \end{matrix}| \to_{R_{4} \Leftrightarrow R_{7}; R_{5} \Leftrightarrow R_{6}}^{R_{1} \Leftrightarrow R_{10}; R_{2} \Leftrightarrow R_{9}, R_{3} \Leftrightarrow R_{8}} {(- 1)}^{5} |\begin{matrix} (1 - t) & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & (2 - t) & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & (3 - t) & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & (4 - t) & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & (5 - t) & 6 & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & (6 - t) & 7 & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & (7 - t) & 8 & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & (8 - t) & 9 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & (9 - t) & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & (10 - t) \end{matrix}| N o w t h e a b o v e d e t e r m i n a n t c a n b e c a l c u l a t e d b y m u l t i p l y i n g t h e d i a g m o n a l e l e m e n t s, t h u s d e t (B_{t}) = {(- 1)}^{5} (1 - t) (2 - t) (3 - t) (4 - t) (5 - t) (6 - t) (7 - t) (8 - t) (9 - t) (10 - t) d e t (B_{t}) = - (1 - t) (2 - t) (3 - t) (4 - t) (5 - t) (6 - t) (7 - t) (8 - t) (9 - t) (10 - t)$

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