Question

s1 3 2 Consider the matrix X = WNA WNNE 2 4 13 6 9 1 9 6 • Find rref(X) • Find a basis for, and the dimension of, ker(X) and of im(x).

Answer 1

Consider the matrix- 1 2 3 21 X=12 4 1 2 12 4 9 1 2 L3 6 9 6 3 find rref(x) Row - Reduced Eecheleon form- Nnto sol -TNM Apply Row operation- R, - R₂R2 [ 2 3 2 17 Loo - 1 - 1 2 4 9 1 2 +3 6 9 R2-2R, GR, and R3- 3R, E R3 1 2 3 2 17 To 0 - 1 - 1 0 0 3 -3 0 To 0 0 0 ol 3 R₂ + R₂ R3 [1 2 3 2 1 To 0 1 1 - 1 oooo 0 -3. Too ooo -R₂ R2 & -R3 AR OOO

2 3 2 17 To 0 -1 0 o o o Looo ool This is Required sref (x) incircle element is called Pivot element Img (x) is spanned by linearly independent of Coloumn Vector. In which coloumn Pivot element exist Corresponding coloumn in x are linearly independent نر) ه های Thue, - - Basis far It dimension of Lng (x) = 3.

Ker (X) = ?X.Y=0 ,YFx2 123 24 175 1 2 3 2 111 XY=11 2 4 1 21/X2 12 4 9 ! 2123 3 6 9 6 3 ه 714 1 2 3 2 17 0 0 1 -1 / 22 100 oo 11/13 = 0 0 0 0 of lar X4 x + 2x2 + 3x₃ +224 +25= 0 X3-14 +25=0 75=0 Three equation five variable = two free variable. ( two LI Vector) Firstl so x = 1, X2=0. then, 1+ 3x3+224 +25=0 X3=14, x5=0

3x3 + 2X2 = 1 5x3 = -1 x3 = -1 =24 I 17 y= 0 1-1 og so o llovi 5 LoJ Now, het second Vector X =0 12=1 & x50 x3=14 0 + 2 + 3x3+2 523 = -2 X ! x3 ! Lion ON NVO 2151 Thurs, Basss of Keri(x) ollow ONN vo dim ker(x) = 21