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)=

The rank of a matrix is the maximum number of linearly independent rows or columns. In this case, looking at the row-reduced echelon form (RREF), we can see that there are 4 non-zero rows. Therefore, the rank of A is 4.

The nullity of a matrix is the dimension of the null space, which is the set of all vectors that when multiplied by the matrix give the zero vector. It can also be found by subtracting the rank of the matrix from the number of columns. In this case, the matrix A is a 5x5 matrix and the rank is 4, so the nullity of A is 5 - 4 = 1.

The kernel or nullspace of a matrix A is the set of all vectors x such that Ax = 0. To find a basis for the kernel, we need to find a solution to this equation. Looking at the RREF, we can see that the fifth column does not have a leading 1, meaning that the corresponding variable is free. We can set this variable to 1 and the others to 0 to find a vector in the nullspace. Therefore, a basis for the kernel of A is {<0, 0, 0, 0, 1>}.