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…

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter6: Matrices And Determinants

Section: Chapter Questions

Problem 11CC

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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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