1.Let the null space of matrix G and the null space of matrix H_be orthogonal complements. Thenonly one of the following statement can be true:N(G) = N(H)b. N(G")=N(H") |© N()=C(H")a.d.N(G)= C(H)If the column space of a 3×3 matrix consists of all vectors b=|b2 such that b, +b2 = 3b3 thenone of the following set of vectors forms a basis for the left null space of that matrix:Vaリ(6)33b.)and3d.3с.а.and3If the vector b=|3| is projected onto the line defined by the intersection of the two planesx+ y+z =Q and x+2y+z=0, then the projection vector is equal to11d.b.a.-14The vector-1 is the orthogonal complement of the space spanned by the vectors7[2-40.14b.C.33d.'a.7C(A"), where A=2 916 5 c(AT)One of the following vectors belongs to-3b-24-2a--34-10A is an mxn matrix with rank r and the system A'y= b has exactly one solution for some b(but not for every possible b ), thenb.C.m>nd.m2nn=r

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Let the null space of matrix G and the null space of matrix H be orthogonal complements. Then only one of the following statement can be true: N(G) = N(H) 6. N(G")=N(H") N(G)=C(H") d. N(G)= C(H) %3D a. If the column space of a 3x3 matrix consists of all vectors b=|b2 | such that b, +b, = 3b, then b3 one of the following set of vectors forms a basis for the left null space of that matrix: (9)* N 3 3 (b. 3 and 3 d. 3 с. a. 1 and 3 -3 -3

Let the null space of matrix G and the null space of matrix H be orthogonal complements. Then only one of the following statement can be true: N(G) = N(H) 6. N(G")=N(H") N(G)=C(H") d. N(G)= C(H) %3D a. If the column space of a 3x3 matrix consists of all vectors b=|b2 | such that b, +b, = 3b, then b3 one of the following set of vectors forms a basis for the left null space of that matrix: (9)* N 3 3 (b. 3 and 3 d. 3 с. a. 1 and 3 -3 -3

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.2: Linear Independence, Basis, And Dimension

Problem 3AEXP

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