Under LU factorisation of

[
8

2

2

4

4

2

0

1

3
]

write down the value of
𝑈
23
(correct to 1 decimal place)

Question

Under LU factorisation of

[
8
	
2
	
2

4
	
4
	
2

0
	
1
	
3
]

write down the value of 
𝑈
23
 (correct to 1 decimal place)

BlackTom AI · Accepted Answer

Use Doolittle (L has 1s on the diagonal) so A = LU with
A = [8  2  2; 4  4  2; 0  1  3]. The relevant concepts: in Doolittle we set U’s first row equal to the first row of A, compute L multipliers for column 1, then form the Schur complement for the remaining 2×2 block and continue.

Step-by-step calculation with all intermediate values:

1. First row of U:
U11 = 8, U12 = 2, U13 = 2.

2. Compute L multipliers for column 1:
L21 = A21 / U11 = 4 / 8 = 0.5
L31 = A31 / U11 = 0 / 8 = 0

3. Form the second row of U by subtracting L21 times the first row of U from the second row of A:
U22 = A22 − L21·U12 = 4 − 0.5·2 = 4 − 1 = 3
U23 = A23 − L21·U13 = 2 − 0.5·2 = 2 − 1 = 1

(These are the direct Doolittle updates; U23 is obtained at this step.)

4. For completeness, the remaining L and U entries would be:
L32 = (A32 − L31·U12) / U22 = (1 − 0·2) / 3 = 1/3
U33 = A33 − L31·U13 − L32·U23 = 3 − 0 − (1/3)·1 = 8/3 ≈ 2.6667

Why U23 = 1 is correct:
The value U23 arises from eliminating the first column below the pivot 8; subtracting 0.5 times the first row from the second row changes the (2,3) entry from 2 to 1 exactly, so U23 = 1 (to one decimal place 1.0).

Other options: not applicable here (no multiple-choice list provided).

Conclusion: U23 = 1, which matches the provided answer.

Under LU factorisation of [ 8 2 2 4 4 2 0 1 3 ] write down the value of 𝑈 23 (correct to 1 decimal place)数值题

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