Question

-1 3 A= = (271 0 (a) Find the nullspace of A. (b) Do the columns of Aform a spanning set for R2? Clearly explain why or why not.

Answer 1

If you have any doubts in the solution please ask me in comments...here i use reduced row echlon form of A

A= (2 - 3 first to find Null space of A. @ Null spae (A) = { x: { x: De IR3 Ax = 0 }, Now reduced Now echion form of A. first we found 3 RI R2 3 A= - (2, 3 ) aw -6 R₂ R2-2R, 3 R2 > 13 R2 R, → R, TR2 3 ló ) -2) 0 0 -2 - - 2 1 0 300 So Deduced echion is form of A ló 2) -2 NOω Let XEIR² 26 Nov (A), -1 3 21 Ax = 0 0 ow = X2 23 (8). 2 --0 By using reduced row echion form we have the system o will be . (69_))(1)-(3) = 2, +383 = 0, X2 - 2xz O + > X; = -133, R2 = 223 21 183 -> +223 +2 x2 13 1 as so is Basis for Null space of A 2 1

> Null space of A = Span {el] Answer (b ) NOW we claim that column of A form a spanning set for IR? is O 2 ) -2 Because we have reduced 30W echion form of A Obich has two pinot (which is circled) 0 0 so corrosponding pivot column the Ist 4 2nd column Lineasly independent it foom a Basis for CO(A). ane apan [:). (61 CUT(A) = 2. Now dimension of Col(A) = is subspace of IR2. as que know that COICA) (there are 2 rows in A). ܝ ' IR ² has dimension 2, so any subspace of IR² has at- most 2 dimension is which IR2. col(A) subespace equal to = > COICA) has dimension 2; of 1R2 is Basis also → C01(A) Basis for IR? a spanning set ex for IR² collemn (A) foom IR? - span {[?], [1] CO! (A) = form a spanning set for IR². column of A Hence proove