System of Linear Equations - Sherman Morrison Method
A = [ 100 44.7043 18.2617 − 47.4786 10.9048 36.4116 20.3398 − 35.4335 2.2131 ] A = \begin{bmatrix} 100 & 44.7043 & 18.2617 \\ -47.4786 & 10.9048 & 36.4116 \\ 20.3398 & -35.4335 & 2.2131 \end{bmatrix} A = 100 − 47.4786 20.3398 44.7043 10.9048 − 35.4335 18.2617 36.4116 2.2131 Find inverse of A.
Explanation
Explanation
Explanation
Explanation
A − 1 = [ − 0.0030065 − 0.0030065 − 0.0030065 0.027422 0.027422 0.027422 0.0085493 0.0085493 0.0085493 ] u = [ 2 − 3 3 ] v = [ − 1 − 2 − 3 ] A^{-1} = \begin{bmatrix} -0.0030065 & -0.0030065 & -0.0030065 \\ 0.027422 & 0.027422 & 0.027422 \\ 0.0085493 & 0.0085493 & 0.0085493 \end{bmatrix} u = \begin{bmatrix} 2\\ -3\\ 3 \end{bmatrix} v = \begin{bmatrix} -1\\ -2\\ -3 \end{bmatrix} A − 1 = − 0.0030065 0.027422 0.0085493 − 0.0030065 0.027422 0.0085493 − 0.0030065 0.027422 0.0085493 u = 2 − 3 3 v = − 1 − 2 − 3 Given update vetors u and v, compute the factors as follows,
P = A − 1 × u P = A^{-1} × u P = A − 1 × u Q = v T × A − 1 Q = v^{T} × A^{-1} Q = v T × A − 1 denominator = [ 1 ] + Q × u \text{denominator} = [1] + Q × u denominator = [ 1 ] + Q × u B = P × v T × A − 1 B = P × v^T × A^{-1} B = P × v T × A − 1 Find B.
Explanation
Explanation
Explanation
Explanation
Submit Quiz
