\\section{Dead ends in PageRank computations (25 points)} Let the {hem matrix of the Web} M be an $n$-by-$n$ matrix, where $n$ is the number of Web pages. The entry $m_{ij}$ in row $1$ and column $j$ is 0, unless there is an arc from node (page) $j$ to node $1$. In that case, the value of $m_{ijl$ is $1/k$, where $k$ is the number of arcs (links) out of node $j$. Notice that if node $j$ has $k>0$ arcs out, then column $j$ has $k$ values of $1/k$ and the rest 0's. If node $j$ is a {\\em dead end} (i. e., it has zero arcs out ), then column $j$ is all 0's. \\def\\bfr{{\\bf }} Let $\\bfr = [r_1, r_2, \\ldots, r_n]^T$ be (an estimate of ) the PageRank vector; that is, $r_i$ is the estimate of the PageRank of node $1$. Define $w(\\bfr )$ to be the sum of the components of $\\bfr$; that is $w(\\bfr ) = \\sum_{i=ijn _i$. In one iteration of the PageRank algorithm, we compute the next estimate $\\bfr'$ of the PageRank as: $\\bfr' = M\\bfr$. Specifically, for each $1$ we compute $r'_i = \\sum_{j=1}^n M_{ij} r_j$. Define $w(\\bfr' )$ to be the sum of components of $r'S; that is $w(\\bfr' ) = \\sum_{i=1}^n r_i's. You may use D (the set of dead nodes) in your equation. `subquestion{(a) [6pts]} Suppose the Web has no dead ends. Prove that $w(\\bfr' ) = w( \\bfr)$. \\subquestion{(b) [9pts]} Suppose there are still no dead ends, but we use a teleportation probability of $1-\\beta$ where we teleport to a random node $(0