Partial preview of the text
Download some questions about matrices and more Exercises Mathematics in PDF only on Docsity!
1-4 2 EXAMPLE 1 Compute det 4, where A = 3 8 3| -1 7 0 SOLUTION The strategy is to reduce A to echelon form and then to use the fact that the determinant of a triangular matrix is the product of the diagonal entries. The first hwo row replacements in column 1 do not change the determinant: 42 deta 9-5 7 0 1-4 2 dtad=—-|0 3 2) = -(1)(3-5)=15 . 0 0-5 [Determine whether the sets in Exercises 1-8 are bases for 2 JOF the sets that are not bases, determine which ones are linearly lindependent and which ones span 3. Justify your answers. 1 0 0 2 2 8 afijjof}a] 4 )-1f).f-3).} 5 0 0 1 1 2 4 zero vector exist, theyare row reduce the given vectors, it leads to the Identity matrix 3 pivots->yes they form basis for R3 not linearly independent For each subspace in Exercises 1-8, (a) find a basis for the subspace, and (b) state the dimension. 1 2 s—2t H =Span{v,,v;}, where v=) 1) and v=) 1 1 set jis,tinE 0 3 3t Since v, and v; are not multiples of each other, ty, Hence the dimension of H is 2. } is linearly independent and is thus a basis for //, In Exercises 13 and 14, assume that A is row equivalent to B. Find bases for Nul 4 and Col A. [ L 2 3 -4 3 | 1 2 0 2 8 A= ,| 1st ard and Sth are pivot a [3 4-3 10 | columns. so basis for Col A is: 3.60 6 9 - ip ayfe 12 0 2 5 lattes B= oO OQ 3 =—6 3 ah'h3)) 9 o oO @ 0 -7 3 oll9 0 0 0 0 0 To find a basis for Nul 4, we find the general solution of Ax = 0 in terms of the free variables, mentally completing the row reduction of & to get: 2x, —4x,, x, =(7/5)X4, X¢=0, with x, and x, free. So 2] “t and a basis for Nul 4 is 1 o -2] [ -4]] | ol+x,| 7/5], 1 0} | | of 1 o),) 7/5}. Lo} a o}) ai] o}| of} In Exercises 11 and 12, find the dimension of the subspace spanned by the given vectors. 1 3 2 5 flo 3 2 | Oj,) 1 j,) -lj,} 2] 0 1 4 33 2 1 1 2\] lo 0 0 «© There are wo pivot columns, so the dimension of Col4 is 2, [Determine the dimensions of Nul A and Col A for the matrices shown in Exercises 13-18. There are 3 pivot columns, so dimension of Col A=3. 1 || There are 2col without pivot, so the equation Ax = 0 has 2 free var. Thus the dimension of Nul A=2. [EXAMPLE 2 Find bases for the row space, the column space, and the null space of he matrix i A 2 -5 8 0 -17 1 3-5 1 3 A 1 3-5 15 o 1-2 2 -7 es) 3 i -19 7 1], ~}0 0 6 =-4 20 17-18 5 -3 o 0 09 0 © In Exercises 1=—4, assume that the matrix A is row equi Without calculations, list rank A and dim Nu! for Col A, Row A, and Nul A. 1 —-4 9 -7 HAs [-: 2 =4 ‘|. 3 -6 10 7 1 Oo -1 5 B=|0 -2 5 -6 oO oO oO 0 A basis for Row 4 is the pivot rows of B: {(1,0,-1,5),(0,-2,5,-6)}. reduced echelon form S/2)x,—3xy (; Oo =r xX, and x, free. Thus a basis for Nul 4 is A~ oO 5/2 [1 5/2 1 O)and[-5 -3 0 1) EXAMPLE 1. uaaa[i “afte [71 ante = [ 7) te imanes or wana ‘AY iS just 2¥. So 4 only The basic Idea is here . 6 6 3 5 Ces [_f]entv= [af anemevaeee 1 6 6 -24 6 w=[5 SI] 5]=[%]-—[s]--« 6 3 9 3 s ‘l[2]-[a]+-[3] ‘Thus u is an cigenvectorcorresponding to an eigenvalue (—4), but v isnot an eigenvector ‘of A, because Av is not a multiple of ¥ . 2pivot so, dim (Col A) 2 free so, dim (Nul A) =2 A basis for Col 4 3] -] x = 3m. ¥ under multiph by A are shown in Fig. 1. “stretches,” or dilates, v. EXAMPLE 2 Let vectors of A? SOLUTION av= Show that 7 is corresponding eigenvectors. eigenvalue of [In Exercises 15—18, finda basis for the space spanned by the given vectors, v1,....¥s. "space spanned by vi,v2... is Col(A)" AMLIB LTTE Fs pean spanned by the given vectors is same as 14 103 1 2 f1 0 30 4 nro) Oo 14 3 1) jo t 4 0 =s of} | a| fst} 3 2 1 8 6) 0 0 oO 1 -2) call] 2)l-e|f 23 6 7 9J (0 0 00 0 2) \-3 7h SOLUTION The scalar 7 is an and only if the equation AX = 7x a has a nontrivial solution. But (1) is equivalent to Ax — 7x = 0, or (A-7P)x = 0 Q) ‘To solve this homogeneous equation, form the matrix 1 6]_[7 0 -6 6 cor LE (4 4] The columns of A —7/ are obviously linearly dependent, so (2) has nontrivial solu-| Thus 7 is an eigenvalue of A. Tolfind the corresponding eigenvectors. Use row) 6 6 0 1-1 0 5-5 o]~[o 0 0 a 1 ‘The general solution has the form af i } Each vector of this form with xz # 0 isan 
