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Given $$A = \pmatrix{2 & 0 & -1 & \\ -1 & 1 & 1 \\ -1 & 0 & 1}$$ and $$A^{-1} = \pmatrix{1 & 0 & 1 & \\ 0 & 1 & -1 \\ 1 & 0 & 2}$$ I want to use the Sherman-Morrison Formula to determine $A^{-1}$ if we update $A$ so that the $a_{32}$ element is $2$. In other words, now $$A = \pmatrix{2 & 0 & -1 & \\ -1 & 1 & 1 \\ -1 & 2 & 1}$$

I've seen how to do this for $2\times2$ matrices, but I cannot seem to figure out how to do this for $3 \times 3$ matrices.

Note: I believe the answer is the following: $$\pmatrix{1 & 2 & -1 & \\ 0 & -1 & 1 \\ 1 & 4 & -2}$$

peterh
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GaMbiTaaaa
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2 Answers2

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To update $A$ such that $a_{3,2}$ is now $2$, you could consider vectors $u=(0,0,2)^T$ and $v=(0,1,0)^T$ as $$A+uv^T=A_{updated}$$

and substitute these vectors into the Sherman-Morrison formula, given $A$ and $A^{-1}$:- $$(A+uv^T)^{-1}=A^{-1}+\frac{A^{-1}uv^TA^{-1}}{1+v^TA^{-1}u}$$

Alijah Ahmed
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you want to find the inverse of $A + 2e_3e_2^T?$ we will use the facts that $$(I + ab^T)^{-1} = I - \frac{ab^T}{1 + a^Tb}, (A+ab^T)^{-1} = A^{-1} - \frac{A^{-1}ab^TA^{-1}}{1+b^TA^{-1}a.}$$

we will take $$a = 2e_3, b =e_2, A^{-1}a = 2(1,-1,2)^T, b^TA^{-1} = (0,1,-1), b^TA^{-1}a=-2$$ putting all these together gets you $$(A + ab^T)^{-1} = A^{-1} +2 \pmatrix{1\\-1\\2}\pmatrix{0&1&-1} =\pmatrix{1&0&1\\0&1&-1\\1&0&2} +2\pmatrix{0&1&-1\\0&-1&1\\0&2&-2} = \pmatrix{1&2&-1\\0&-1&1\\1&4&-2} $$

abel
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