Consequently the set of vectors 1 1 5 4 1 1 is an

This preview shows page 6 - 8 out of 8 pages.

00910.0pointsThe matrixA=213042102is invertible.True or False?10.0pointsIfAis a 4×5 matrix, thendim(Col(A)) + dim(Nul(A)) = 5.True or False?
2.FALSE
01110.0pointsIfAis a 4×5 matrix, thendim(Col(A)) + dim(Nul(A)) = 5.True or False?
Version 012 – EXAM 03 – gilbert – (53415)7TRUE.01310.0pointsIf{x1,x2,x3}is a linearly independent setandW= Span{x1,x2,x3},then any orthogonal set{v1,v2,v3}inWisa basis forW.True or False?1.TRUE2.FALSEcorrect
True or False?The three orthogonal vectors must benon-zeroto be a basis for a three-dimensionalsubspace.
Explanation:SetA=264352163,and132352163n×nmatrixAare abasis forRn, thenAis diagonalizable.True or False?
Consequently, the statement isTRUE.01310.0pointsIf{x1,x2,x3}is a linearly independent setandW= Span{x1,x2,x3},then any orthogonal set{v1,v2,v3}inWisa basis forW.True or False?1.TRUE2.FALSEcorrect
Explanation:The three orthogonal vectors must benon-zeroto be a basis for a three-dimensionalsubspace.
01410.0pointsIf eigenvectors of ann×nmatrixAare abasis forRn, thenAis diagonalizable.True or False?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture