Question

W is an n x n matrix.

T F if the rows of W are linearly independent, then the detW is not equal to zero

T F if the detW is not equal to zero, then the rank of W is equal to n.

T F if Wx = V has a unique solution for every V in Rn, then the detW is not equal to zero.

T F if null(W) = {0}, then the detW is equal to zero.

T F if det(W - λi) = 0, then Wx = 0 has infinitely many solutions

T F if λ = 0 is an eigenvalue of W, then W is invertible

Answer 1

solution: Wis an nxn matrix Consider a 2x2 matria in genral W= [ 34 x -() Now Here clearly rows are linearly independent 1A)= 15214-6 =-2 +0 I Alto Wo This result will hold for all nan matrices. Thue it is true that if the rows of W are linecarly independent then det W is not equal to zero. consider a 1 3x3 matrix WE 2 37 4 5 6 7 8 9 373 Here clearly rows are linearly inde pendent Rank N= 3 (When n=3) Also from (i) Rank Iw=2} When=2) Thus when we will take matrix . Rank W=n ( When order of W-is non) then Thus it is true that if the det lN is not equal to o. the rank of W is equal to n. a non then if I W-dil=0 can 1MX=0 has infinitly many solution it is True Because for same eigen value you have different different many ergen Vecton that will led you. infinitely many solutions. if 1:0 * is an eigen value of n. that product of eigen values = cet W. We know

det w=0 11 lie. din 12 ..... do det W suppose dıo (given) Ox 12 x 13X.. Xans cet w det W=0 Thus W is and can't be invertible. ". It is false ifA=0 is an eigen value of al, then ln is invertible. Rank W=n las we have concluded ) ie. glinan TW140 if IW=o then plu). can't n. Thus it is true if INX-V has a unique solution for every v in Irn , then the det n is not equal to zero. it is only possible when