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How do I solve this?

How do I solve this?-example-1
User Machisuji
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1 Answer

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(i) Each of u, v, and w are vectors in R, so they each have size n × 1 (i.e. n rows and 1 column). So u and v both have size n × 1, while wᵀ has size 1 × n.

M is an n × n matrix, so the matrix A has been partitioned into the blocks


A=\begin{pmatrix}M_(n* n)&\mathbf u_(n* 1)\\\mathbf w^\top_(1* n)&\alpha\end{pmatrix}

where α is a scalar with size 1 × 1. So A has size (n + 1) × (n + 1).

(ii) Multiplying both sides (on the left is the only sensible way) by the given matrix gives


\begin{pmatrix}M^(-1)&\mathbf 0\\-\mathbf w^\top M^(-1)&1\end{pmatrix}\begin{pmatrix}M&\mathbf u\\\mathbf w^\top&\alpha\end{pmatrix}\begin{pmatrix}\mathbf x\\x_(n+1)\end{pmatrix}=\begin{pmatrix}M^(-1)&\mathbf 0\\-\mathbf w^\top M^(-1)&1\end{pmatrix}\begin{pmatrix}\mathbf v\\v_(n+1)\end{pmatrix}


\begin{pmatrix}M^(-1)M&M^(-1)\mathbf u\\-\mathbf w^\top M^(-1)M+\mathbf w^\top&-\mathbf w^\top M^(-1)\mathbf u+\alpha\end{pmatrix}\begin{pmatrix}\mathbf x\\x_(n+1)\end{pmatrix}=\begin{pmatrix}M^(-1)&\mathbf 0\\-\mathbf w^\top M^(-1)&1\end{pmatrix}\begin{pmatrix}\mathbf v\\v_(n+1)\end{pmatrix}

and of course M ⁻¹ M = I (the identity matrix), so

-wM ⁻¹ M + wᵀ = -wᵀ + wᵀ = 0ᵀ (the zero vector transposed)

(iii) Simplifying the system further gives


\begin{pmatrix}I&M^(-1)\mathbf u\\\mathbf 0^\top&-\mathbf w^\top M^(-1)\mathbf u+\alpha\end{pmatrix}\begin{pmatrix}\mathbf x\\x_(n+1)\end{pmatrix}=\begin{pmatrix}M^(-1)&\mathbf 0\\-\mathbf w^\top M^(-1)&1\end{pmatrix}\begin{pmatrix}\mathbf v\\v_(n+1)\end{pmatrix}


\begin{pmatrix}\mathbf x+x_(n+1)M^(-1)\mathbf u\\(\alpha-\mathbf w^\top M^(-1)\mathbf u)x_(n+1)\end{pmatrix}=\begin{pmatrix}M^(-1)\mathbf v\\-\mathbf w^\top M^(-1)\mathbf v+v_(n+1)\end{pmatrix}

So now, setting y = M ⁻¹u and z = M ⁻¹ v gives


\begin{pmatrix}\mathbf x+x_(n+1)\mathbf y\\(\alpha-\mathbf w^\top\mathbf y)x_(n+1)\end{pmatrix}=\begin{pmatrix}\mathbf z\\-\mathbf w^\top \mathbf z+v_(n+1)\end{pmatrix}

Given that α - wy ≠ 0, it follows that


x_(n+1)=(v_(n+1)-\mathbf w^\top\mathbf z)/(\alpha-\mathbf w^\top\mathbf y)

(iv) Combining the result from (iii) with the first row gives


\mathbf x+x_(n+1)\mathbf y=\mathbf z


\mathbf x=\mathbf z-x_(n+1)\mathbf y


\mathbf x=\mathbf z-(v_(n+1)-\mathbf w^\top\mathbf z)/(\alpha-\mathbf w^\top\mathbf y)\mathbf y

User Est
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