need the correct and unique solves asap plz. do not copy from chegg
![lai.j||T;|< (Eitj lai,j)|T:| using (2).<br>Prove that diagonally dominated matrices are always invertible.<br>Now, what is a diagonally dominated matrix? It is a square matrix that has each of it's diagonal<br>values larger in magnitude than all the other values in the rows combined.<br>To be mathematically precise, for every diagonal value a;i,<br>lai,i| > Eitj lai.jl .(1)<br>lai,i| > Ja;,1[+ |ai,2|+ ... + |ai,i-1|+ |ai,i+1|+ ....+ |ai,n|<br>For example,<br>3 1<br>5 2<br>1<br>1<br>1<br>2<br>-3<br>1<br>3<br>Hints:<br>1<br>1<br>4<br>• Remember that, a matrix A is invertible if the only solution to Ar = 0 is the 0 vector, i.e.<br>x = 0.<br>• Suppose, for a DDM (diagonally dominated matrix), there exists a non-zero solution to<br>Ax = 0. Since x =<br>Let's say, ri has the greatest magnitude here which means |xi| > |x;| for all j# i ...(2).<br>• Since, Ax = 0, we must have,<br>ai,1X1 + ai,2x2 + · · ·+ a¿.n²n = 0 for the i'th row.<br>In other words,<br>ai,1x1 + ai,2x2 + .+ ai,i-1Li-1+ai,i+1Xi+1+ • · · + ai,nIn = -ai,iXi<br>or,<br>lai,1x1 + ai,2x2 + ...+ ai,i-1x;-1+ a¿,i+1Xi+1+.+ ai,nXn| = |ai,ix¡|<br>TEitj ai,j®j| = |a;,i||x||<br>But we will prove that, this is not possible when x is on-zero.<br>So, we have proved that, |Eit; aijX;|< |ai,i¤i|<br>Explanation:<br>TEitj ai,jªj| <Ei+j\ai.jx;| because |r+ y| < |x| + \y] for all real numbers r, y, z.<br>lai.jaj| = Ei+j |ai,j||x;| because |xy| = |x||y| for all real r, y.<br>(Eitj laij|)|xi| < |a;,i||r| using (1)<br>](https://s3.us-east-1.amazonaws.com/storage.unifolks.com/qimg-008/008_yvnmxz0-ryyyyxis.jpeg)
Extracted text: lai.j||T;|< (eitj="" lai,j)|t:|="" using="" (2).="" prove="" that="" diagonally="" dominated="" matrices="" are="" always="" invertible.="" now,="" what="" is="" a="" diagonally="" dominated="" matrix?="" it="" is="" a="" square="" matrix="" that="" has="" each="" of="" it's="" diagonal="" values="" larger="" in="" magnitude="" than="" all="" the="" other="" values="" in="" the="" rows="" combined.="" to="" be="" mathematically="" precise,="" for="" every="" diagonal="" value="" a;i,="" lai,i|=""> Eitj lai.jl .(1) lai,i| > Ja;,1[+ |ai,2|+ ... + |ai,i-1|+ |ai,i+1|+ ....+ |ai,n| For example, 3 1 5 2 1 1 1 2 -3 1 3 Hints: 1 1 4 • Remember that, a matrix A is invertible if the only solution to Ar = 0 is the 0 vector, i.e. x = 0. • Suppose, for a DDM (diagonally dominated matrix), there exists a non-zero solution to Ax = 0. Since x = Let's say, ri has the greatest magnitude here which means |xi| > |x;| for all j# i ...(2). • Since, Ax = 0, we must have, ai,1X1 + ai,2x2 + · · ·+ a¿.n²n = 0 for the i'th row. In other words, ai,1x1 + ai,2x2 + .+ ai,i-1Li-1+ai,i+1Xi+1+ • · · + ai,nIn = -ai,iXi or, lai,1x1 + ai,2x2 + ...+ ai,i-1x;-1+ a¿,i+1Xi+1+.+ ai,nXn| = |ai,ix¡| TEitj ai,j®j| = |a;,i||x|| But we will prove that, this is not possible when x is on-zero. So, we have proved that, |Eit; aijX;|< |ai,i¤i|="" explanation:="" teitj="" ai,jªj|="">
< |x|="" +="" \y]="" for="" all="" real="" numbers="" r,="" y,="" z.="" lai.jaj|="Ei+j" |ai,j||x;|="" because="" |xy|="|x||y|" for="" all="" real="" r,="" y.="" (eitj="" laij|)|xi|="">< |a;,i||r| using (1) |a;,i||r|="" using="">