Please generalize this code to work for a matrix of any dimension:

def isSingular$($A$)$ :

B $=$ np$.$array$($A$,$dtype$=$np$.$float$\_)$ # Lets make a copy as we are going to change this matrix

try:

alterRowZero$($B$)$

print$($B$)$

alterRowOne$($B$)$

print$($B$)$

alterRowTwo$($B$)$

print$($B$)$

alterRowThree$($B$)$

print$($B$)$

except ItIsSingular:

return True

return False

#this is our error class. no need to make any changes here

class ItIsSingular$($Exception$)$: pass

#Remember that for the first row, all we need is the first element to be $1.$

#We can do it by dividing the whole row by the value of the element i$.$e$.,$ A$[0,0]$

#However if A$[0,0]$ is zero, then we are in trouble.. so need to test for that.

#if that is the case we can add the lower row one by one to make sure we have a non zero first element.

# Do not edit this.

def alterRowZero$($A$)$ :

if A$[0,0]==0$ :

A$[0]=$ A$[0]+$ A$[1]$

if A$[0,0]==0$ :

A$[0]=$ A$[0]+$ A$[2]$

if A$[0,0]==0$ :

A$[0]=$ A$[0]+$ A$[3]$

if A$[0,0]==0$ :

raise ItIsSingular$()$

A$[0]=$ A$[0]/$ A$[0,0]$

return A

#We need to make the element below the diagonal element zero i$.$e$.$ A$[1,0]$

# Next make A$[1,1]=1$

# We'll divide the row by the value of A$[1,1].$

# Do the zero test again

# There is no need to edit this function.

def alterRowOne$($A$)$ :

A$[1]=$ A$[1]-$ A$[1,0]*$ A$[0]$

if A$[1,1]==0$ :

A$[1]=$ A$[1]+$ A$[2]$

A$[1]=$ A$[1]-$ A$[1,0]*$ A$[0]$

if A$[1,1]==0$ :

A$[1]=$ A$[1]+$ A$[3]$

A$[1]=$ A$[1]-$ A$[1,0]*$ A$[0]$

if A$[1,1]==0$ :

raise ItIsSingular$()$

A$[1]=$ A$[1]/$ A$[1,1]$

return A

# Complete this function

def alterRowTwo$($A$)$ :

# Make the required elements in row two to zero $($hint: there are $2$ of them$).$

A$[2]=$ A$[2]-$ A$[2,0]*$ A$[0]$

A$[2]=$ A$[2]-$ A$[2,1]*$ A$[1]$

# zero test

if A$[2,2]==0$:

A$[2]=$ A$[2]+$ A$[3]$

A$[2]=$ A$[2]-$ A$[2,0]*$ A$[0]$

# make sub diagonal elements zero

if A$[2,2]==0$:

raise ItIsSingular$()$

# Finally set the diagonal element to one by dividing the whole row by that element.

A$[2]=$ A$[2]/$ A$[2,2]$

return A

# You should also complete this function

# Follow the instructions inside the function at each comment.

def alterRowThree$($A$)$ :

# Insert code below to set the sub$-$diagonal elements of row three to zero.

#Setting elements in row three to zero

A$[3]=$ A$[3]-$ A$[3,0]*$ A$[0]$

A$[3]=$ A$[3]-$ A$[3,1]*$ A$[1]$

A$[3]=$ A$[3]-$ A$[3,2]*$ A$[2]$

#sub$-$diagonal elements zero

if A$[3,3]==0$:

raise ItIsSingular$()$

#Sub$-$diagonal elements to one

A$[3]=$ A$[3]/$ A$[3,3]$

return A