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1. Define KnK_{n} to be the n×nn \times n matrix with 2 on the main diagonal, -1 ...
Mar 18, 2024
1. Define KnK_{n} to be the n×nn \times n matrix with 2 on the main diagonal, -1 on the subdiagonal and superdiagonal, and zeros elsewhere. K4K_{4} is shown below: K4=(2amp;−1amp;0amp;0−1amp;2amp;−1amp;00amp;−1amp;2amp;−10amp;0amp;−1amp;2) K_{4}=\left(\begin{array}{cccc} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array}\right) (a) By performing three row replacement operations, reduce K4K_{4} to echelon form. [1] (b) Without using the cofactor formula, calculate det⁡(K4)\operatorname{det}\left(K_{4}\right). [1] (c) Using the cofactor formula, write an expression for det⁡(Kn)\operatorname{det}\left(K_{n}\right) in terms of det⁡(Kn−1)\operatorname{det}\left(K_{n-1}\right) and det⁡(Kn−2)\operatorname{det}\left(K_{n-2}\right). [2] (d) Hence calculate the value of det⁡(Kn)\operatorname{det}\left(K_{n}\right). [2]
Solution by Steps
step 1
To reduce K4 K_4 to echelon form, we perform the following row operations: R2 = R2 + R1, R3 = R3 + R2, and R4 = R4 + R3
step 2
After performing the row operations, the matrix K4 K_4 becomes the identity matrix I4 I_4
step 3
The determinant of K4 K_4 , denoted as det⁡(K4) \operatorname{det}(K_4) , is the product of the diagonal entries of the echelon form
step 4
Since the echelon form is the identity matrix, det⁡(K4)=1×1×1×1=1 \operatorname{det}(K_4) = 1 \times 1 \times 1 \times 1 = 1
Answer
(a) The echelon form of K4 K_4 is I4 I_4 . (b) det⁡(K4)=1 \operatorname{det}(K_4) = 1 .
Key Concept
Row reduction and determinant calculation of a matrix
Explanation
The determinant of a triangular matrix, including the identity matrix, is the product of its diagonal entries.
step 1
To express det⁡(Kn) \operatorname{det}(K_n) in terms of det⁡(Kn−1) \operatorname{det}(K_{n-1}) and det⁡(Kn−2) \operatorname{det}(K_{n-2}) , we expand Kn K_n along the first row
step 2
The expansion gives det⁡(Kn)=2⋅det⁡(Kn−1)−(−1)⋅(−1)⋅det⁡(Kn−2) \operatorname{det}(K_n) = 2 \cdot \operatorname{det}(K_{n-1}) - (-1) \cdot (-1) \cdot \operatorname{det}(K_{n-2})
step 3
Simplifying the expression, we get det⁡(Kn)=2⋅det⁡(Kn−1)−det⁡(Kn−2) \operatorname{det}(K_n) = 2 \cdot \operatorname{det}(K_{n-1}) - \operatorname{det}(K_{n-2})
step 4
This is a second-order linear homogeneous recurrence relation with constant coefficients
Answer
(c) det⁡(Kn)=2⋅det⁡(Kn−1)−det⁡(Kn−2) \operatorname{det}(K_n) = 2 \cdot \operatorname{det}(K_{n-1}) - \operatorname{det}(K_{n-2}) .
Key Concept
Recurrence relation for the determinant of a tridiagonal matrix
Explanation
The determinant of a tridiagonal matrix can be expressed as a recurrence relation involving the determinants of smaller submatrices.
step 1
To solve the recurrence relation, we use the initial conditions det⁡(K1)=2 \operatorname{det}(K_1) = 2 and det⁡(K2)=3 \operatorname{det}(K_2) = 3
step 2
We then iteratively apply the recurrence relation to find det⁡(Kn) \operatorname{det}(K_n) for n=3,4,… n = 3, 4, \ldots
step 3
For n=3 n = 3 , det⁡(K3)=2⋅det⁡(K2)−det⁡(K1)=2⋅3−2=4 \operatorname{det}(K_3) = 2 \cdot \operatorname{det}(K_2) - \operatorname{det}(K_1) = 2 \cdot 3 - 2 = 4
step 4
For n=4 n = 4 , det⁡(K4)=2⋅det⁡(K3)−det⁡(K2)=2⋅4−3=5 \operatorname{det}(K_4) = 2 \cdot \operatorname{det}(K_3) - \operatorname{det}(K_2) = 2 \cdot 4 - 3 = 5
Answer
(d) det⁡(Kn) \operatorname{det}(K_n) for n=1,2,3,4 n = 1, 2, 3, 4 are 2, 3, 4, and 5 respectively.
Key Concept
Solving a recurrence relation
Explanation
The values of det⁡(Kn) \operatorname{det}(K_n) can be determined by iteratively applying the recurrence relation starting from known initial conditions.
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