Transpose ax+b37=d𝑎𝑥+𝑏37=𝑑, x𝑥 as a subject.

Make a the subject of the formula:𝑦=𝑥𝑎𝑡y=xat

Make x the subject of the formula:𝑏=𝑦𝑥𝑎b=yxa

Make t the subject of the formula:𝑓=𝑝𝑡𝑗f=ptj

A𝐴 is a 5×55×5 matrix with columns given by the vectors a1,a2,a3,a4,a5𝑎1,𝑎2,𝑎3,𝑎4,𝑎5  so,A=(a1  a2  a3  a4  a5)𝐴=(𝑎1  𝑎2  𝑎3  𝑎4  𝑎5) If you need to refer to these column vectors in any answer use a1 for a1𝑎1 , a2 for a2𝑎2 etc. A𝐴  has row-reduced echelon form (RREF)U=𝑈= ⎛⎝⎜⎜⎜⎜⎜⎜1000020000010000010000010⎞⎠⎟⎟⎟⎟⎟⎟1200000100000100000100000 State the rank and nullitty of A𝐴 .rank(A)=rank(𝐴)=   nullity(A)=nullity(𝐴)=     Find a basis for the kernel or nullspace of  A𝐴 ,  ker(A)ker(𝐴) . Your answer must be a set of vectors, that is a sequence of vectors separated by commas enclosed in curly brackets { }.  Vectors can either be entered either using Maple notaation, for example,   < 3, 7 >  for the vector (37)(37) or in terms of the vectors   a1,a2,a3,a4,a5𝑎1,𝑎2,𝑎3,𝑎4,𝑎5  written as  a1, a2, a3, a4, a5.The syntax for typical answers is one of the following two forms: { <1,2,3>, <4,5,6>, <1,1,1> }  or  { a2, a3, a4 } Basis for  ker(A)=ker(𝐴)=

b=at𝑏=𝑎𝑡; solve for t𝑡.