The Correct Answer and Explanation is:
To find the QR factorization of the matrix A, we need to find an orthogonal matrix Q and an upper triangular matrix R such that A = QR. We can achieve this using the Gram-Schmidt process on the columns of A.
Let the columns of A be a₁, a₂, and a₃:
a₁ = [4, 1, 3]
a₂ = [2, 5, -3]
a₃ = [0, -5, 3]
Step 1: Find the columns of Q (an orthonormal basis)
First column:
Let u₁ = a₁ = [4, 1, 3].
The norm is ||u₁|| = √(4² + 1² + 3²) = √(16 + 1 + 9) = √26.
The first column of Q is q₁ = u₁ / ||u₁|| = (1/√26) * [4, 1, 3].
Second column:
First, we find an orthogonal vector u₂.
u₂ = a₂ – proj_u₁(a₂) where proj_u₁(a₂) = ((a₂ ⋅ u₁) / (u₁ ⋅ u₁)) * u₁.
a₂ ⋅ u₁ = (2)(4) + (5)(1) + (-3)(3) = 8 + 5 – 9 = 4.
u₁ ⋅ u₁ = ||u₁||² = 26.
u₂ = a₂ – (4/26) * u₁ = [2, 5, -3] – (2/13) * [4, 1, 3]
u₂ = [2 – 8/13, 5 – 2/13, -3 – 6/13] = [18/13, 63/13, -45/13].
To simplify, we can use a scaled vector, for example (13/9) * u₂ = [2, 7, -5]. Let’s call this simplified vector u₂’.
The norm is ||u₂’|| = √(2² + 7² + (-5)²) = √(4 + 49 + 25) = √78.
The second column of Q is q₂ = u₂’ / ||u₂’|| = (1/√78) * [2, 7, -5].
Third column:
First, we find an orthogonal vector u₃.
u₃ = a₃ – proj_u₁(a₃) – proj_u’₂(a₃).
a₃ ⋅ u₁ = (0)(4) + (-5)(1) + (3)(3) = -5 + 9 = 4.
a₃ ⋅ u₂’ = (0)(2) + (-5)(7) + (3)(-5) = -35 – 15 = -50.
u₃ = a₃ – (4/26)u₁ – (-50/78)u₂’ = [0,-5,3] – (2/13)[4,1,3] + (25/39)[2,7,-5].
u₃ = [0-24/39+50/39, -195/39-6/39+175/39, 117/39-18/39-125/39] = [26/39, -26/39, -26/39] = [2/3, -2/3, -2/3].
Again, we can use a scaled vector, 3 * u₃ = [2, -2, -2], or even simpler, [1, -1, -1]. Let’s call this u₃’.
The norm is ||u₃’|| = √(1² + (-1)² + (-1)²) = √3.
The third column of Q is q₃ = u₃’ / ||u₃’|| = (1/√3) * [1, -1, -1].
So, the matrix Q is:
Q = [ 4/√26 2/√78 1/√3 ]
[ 1/√26 7/√78 -1/√3 ]
[ 3/√26 -5/√78 -1/√3 ]
Step 2: Find the matrix R
Since A = QR and Q is orthogonal (QᵀQ = I), we have R = QᵀA. The entries of R are rᵢⱼ = qᵢ ⋅ aⱼ.
r₁₁ = q₁ ⋅ a₁ = ||u₁|| = √26
r₁₂ = q₁ ⋅ a₂ = 4/√26
r₁₃ = q₁ ⋅ a₃ = 4/√26
r₂₁ = q₂ ⋅ a₁ = 0 (by orthogonality)
r₂₂ = q₂ ⋅ a₂ = ((1/√78)[2,7,-5]) ⋅ [2,5,-3] = (4+35+15)/√78 = 54/√78
r₂₃ = q₂ ⋅ a₃ = ((1/√78)[2,7,-5]) ⋅ [0,-5,3] = (-35-15)/√78 = -50/√78
r₃₁ = q₃ ⋅ a₁ = 0 (by orthogonality)
r₃₂ = q₃ ⋅ a₂ = 0 (by orthogonality)
r₃₃ = q₃ ⋅ a₃ = ((1/√3)[1,-1,-1]) ⋅ [0,-5,3] = (5-3)/√3 = 2/√3
So, the matrix R is:
R = [ √26 4/√26 4/√26 ]
[ 0 54/√78 -50/√78 ]
[ 0 0 2/√3 ]
Final Answer:
Q =
[[ 4/sqrt(26), 2/sqrt(78), 1/sqrt(3) ],
[ 1/sqrt(26), 7/sqrt(78), -1/sqrt(3) ],
[ 3/sqrt(26), -5/sqrt(78), -1/sqrt(3) ]]
R =
[[ sqrt(26), 4/sqrt(26), 4/sqrt(26) ],
[ 0, 54/sqrt(78), -50/sqrt(78) ],
[ 0, 0, 2/sqrt(3) ]]thumb_upthumb_down
