Under LU factorisation of

[
8

2

2

4

4

2

0

1

3
]

write down the value of
𝐿
31
(correct to 1 decimal place)

Question

Under LU factorisation of

[
8
	
2
	
2

4
	
4
	
2

0
	
1
	
3
]

write down the value of 
𝐿
31
 (correct to 1 decimal place)

BlackTom AI · Accepted Answer

Because the (3,1) entry of the original matrix is zero, the multiplier L31 used to eliminate the entry below the (1,1) pivot is zero.

Relevant concepts:
- Doolittle LU factorization writes A = LU where L has 1s on the diagonal and the entries below the diagonal are the elimination multipliers.
- The multiplier to eliminate a_ij using pivot a_kk is l_ik = a_ik / a_kk (using the original or updated value in row i before elimination).

Matrix A:
[ 8   2   2
  4   4   2
  0   1   3 ]

Step-by-step calculation:
1. First pivot a11 = 8.
   - l21 = a21 / a11 = 4 / 8 = 0.5.
   - l31 = a31 / a11 = 0 / 8 = 0. (This is the requested L31.)
2. Form the updated rows after eliminating below the first pivot:
   - Row2' = Row2 − l21 * Row1 = [4,4,2] − 0.5*[8,2,2] = [4−4, 4−1, 2−1] = [0, 3, 1].
   - Row3' = Row3 − l31 * Row1 = [0,1,3] − 0*[8,2,2] = [0, 1, 3].
3. Second pivot U22 = 3, then l32 = (updated a32) / U22 = 1 / 3 ≈ 0.333...
4. The L matrix (with ones on diagonal) is
   [1     0     0
    0.5   1     0
    0    0.333  1]
   so L31 = 0.

Why the correct answer is 0:
- L31 is the multiplier to eliminate the (3,1) element using the (1,1) pivot. Since that element is already 0, no elimination is needed and the multiplier is 0.

Why other possible values would be incorrect:
- 0.5 is l21 (row 2’s multiplier), not l31.
- 1/3 (≈0.333) is l32 (multiplier for elimination in column 2), not l31.
- Any nonzero value for L31 would imply a nonzero a31 to eliminate, which is not the case.

Conclusion: L31 = 0, which matches the provided answer.

Under LU factorisation of [ 8 2 2 4 4 2 0 1 3 ] write down the value of 𝐿 31 (correct to 1 decimal place)数值题

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