e) Consider the matrix 2 3 A: 12 2 5 and the Maple session lower down this page to answer the questions below. i) Show that there are innitely many matrices 511 512 513 B = (521 522 523) such that BA 2 12. ii) Show that there is no matrix C such that AC = 13. > with (LinearAlgebra) : > A := | > ; 2 3 1 2 2 5 > M := ; 21210 M : = 3 2 5 0 1 > GaussianElimination (M) ; 2 1 2 1 0 0 1/2 2 -3/2 1> N := ; 2 3 10 0 N : = 1 201 0 2 50 01 > GaussianElimination (N) ; 2 3 1 0 0 0 1/2 -1/2 1 0 0 0 1 -4 1Solution: i) In order for B to be a left inverse of A we need BA = 12, that is 2 3 (511 512 513) 1 2 _ (1 0) _ 1 521 522 523 2 5 0 Each component of this matrix product gives a linear equation in the bij. There are two distinct sets of equations. For b11,b12,b13 the linear equations are 2[311 + 1312 + 2513 = 1 31211 + 2b12 + 5b13 = O and these are represented by the augmented matrix 2 1 2 1 M2_ To reduce M1 and M2 to row echelon form requires the same set of row opera tions, so to save space and time these are combined in the Maple code as 21210 M'(325|01) The Maple code gives the row reduced form of M and from that the row reduced forms of M1 and M2 can be obtained as, 2 1 2 2 1 2 0 and 0 1/2 2 -3/2 0 1/2 2 1 The row echelon forms have no inconsistent row and a non-leading column in the coefficient matrix so there are infinitely many solutions for b11, b12, b13, b21, b22, b23 and hence there are infinitely many matrices B that are left inverses of A.ii) For the matrix to be a right inverse of A, we need 2 3 1 AC = 1 2 (011 012 C13) 2 0 0 C C C 2 5 21 22 23 This is 3 linear equations in 011, 021, 3 linear equations in 012, 622 and 3 linear equations in 013, 023 with corresponding augmented matrices, 2 3 1 2 3 0 2 3 0 1 2 0 , 1 2 1 and 1 2 0 2 5 0 2 5 0 2 5 1 These are combined in the Maple code into a single matrix N which is reduced to row echelon form and from this we nd the row echelon forms for the the 3 augmented matrices above. That is, These are combined in the Maple code into a single matrix N Which is reduced to row echelon form and from this we nd the row echelon forms for the the 3 augmented matrices above. That is, 2 3 1 2 3 0 2 3 0 0 1/2 1/2 , 0 1/2 1 and 0 1/2 0 0 0 1 0 0 4 0 0 1 The last row of each of these row echelon forms corresponds to an inconsistent equation. So there is no solution for 011, 012, 013, 021, 022, 023 and hence no matrix