(i) Each of u, v, and w are vectors in Rⁿ, 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
[tex]A=\begin{pmatrix}M_{n\times n}&\mathbf u_{n\times 1}\\\mathbf w^\top_{1\times n}&\alpha\end{pmatrix}[/tex]
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
[tex]\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}[/tex]
[tex]\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}[/tex]
and of course M ⁻¹ M = I (the identity matrix), so
-wᵀ M ⁻¹ M + wᵀ = -wᵀ + wᵀ = 0ᵀ (the zero vector transposed)
(iii) Simplifying the system further gives
[tex]\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}[/tex]
[tex]\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}[/tex]
So now, setting y = M ⁻¹u and z = M ⁻¹ v gives
[tex]\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}[/tex]
Given that α - wᵀy ≠ 0, it follows that
[tex]x_{n+1}=\dfrac{v_{n+1}-\mathbf w^\top\mathbf z}{\alpha-\mathbf w^\top\mathbf y}[/tex]
(iv) Combining the result from (iii) with the first row gives
[tex]\mathbf x+x_{n+1}\mathbf y=\mathbf z[/tex]
[tex]\mathbf x=\mathbf z-x_{n+1}\mathbf y[/tex]
[tex]\mathbf x=\mathbf z-\dfrac{v_{n+1}-\mathbf w^\top\mathbf z}{\alpha-\mathbf w^\top\mathbf y}\mathbf y[/tex]
