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Augmented matrix - correct answer a+b=c d+e=f [ a b c ; d e f] Row echelon form (REF) - correct answer 1. All nonzero rows are above any rows of all zeros, called 0-rows
- Each leading entry (leftmost nonzero entry) of a row is in a column to the right of the leading entry of the row above it
- All entires in a column below a leading entry are zero Reduced row echelon form - correct answer All the rules for row echelon for plus...
- The leading entry in each nonzero row is 1
- Each leading 1 is the only nonzero entry in its column General solution of a SLE in parametric vector form - correct answer - set free variables equal to s and t -set up constant matrix, s matrix, and t matrix ex. x1-x4=- x2+x3+2x4= x3=s x4=t
x1=t- x2=2-s-2t [x1 ; x2 ; x3 ; x4]= [-1 ; 2 ; 0 ; 0] + s[0 ; -1 ; 1 ; 0] + t[1 ; -2 ; 0 ; 1] Span of vectors in Rn - correct answer The Span{v1, v2, ..., vp} is the set of all linear combinations of vectors v1, v2, ..., vp. In other words, Span{v1, v2, ..., vp} is the collection of all vectors that can be written in the form x1v1 + x2v2 + ... + xpvp where x1, x2, ..., xp are all scalars Linearly independent - correct answer a set of vectors {v1, v2, ..., vp} in Rn where the vector equation x1v1 + x2v2 + ... + xpvp = 0 has only the trivial solution (x1=0, x2=0, & xp=0) Linearly dependent - correct answer a set of vectors {v1, v2, ..., vp} in Rn where there are weights c1, ..., cp not all zero such that the vector equation c1v1 + c2v2 + ... + cpvp = 0 How to determine linear dependence - correct answer -Augment vectors with zero matrix -If there are free variables, the set is linearly dependent -One vector is linearly independent -For two vectors: a set of two vectors is linearly dependent if at least one of the vectors is a multiple of the other. The set is linearly independent if and only if neither of the vectors is a multiple of the other
-If there is a pivot position in every row (no free variables), the equation Ax= 0 has only the trivial solution, and the columns of A are linearly independent Let u and v be two given vectors that are solutions of a SLE Ax = b, where b doesn't =0. Find, if possible, a nontrivial solution of the homogeneous SLE Ax = 0. Say, u = [3, 1, 2, 5]T and v = [1, 1, 2, 1]T. - correct answer A(u-v)= Au-Av= b-b= Solution: u-v Let u and v be two given vectors that are solutions of a SLE Ax = b, where b doesn't =0. Find, if possible, one more solution of Ax=b. Say, u=[3,1,2,5]T and v=[1,1,2,1]T. - correct answer A(cu+dv)=b c+d= c= d=- Solution: 2u-v Invertible Matrix - correct answer An nxn matrix A is said to be invertible if there is an nxn matrix C such that CA=I and AC=I where I=In; the nxn identity matrix.
Inverse C of A - correct answer -When CA=I and AC=I -C is uniquely determined by A, because if B were another inverse of A, then B=BI=B(AC)=(BA)C=IC=C -This unique inverse is denoted by A^-1, so that A^-1A=I and AA^-1=I How to determine if a matrix is invertible - correct answer Let A=[ a b ; c d]... If ad-bc doesn't =0, then A is invertible and A^-1=(1/(ad-bc))*[d -b ; -c a] Is A^-1 invertible? - correct answer If A is an invertible matrix, then A^- is invertible and (A^-1)^-1=A Is AB invertible? - correct answer If A and B are nxn invertible matrices, then so is AB, and the inverse of AB is the product of the inverses of A and B in the reverse order. That is, (AB)^-1=(B^-1)(A^-1) Is A^T invertible? - correct answer If A is an invertible matrix, then so is A^T, and the inverse of A^T is the transpose of A^-1. That is, (A^T)^- 1=(A^-1)^T Elementary matrix - correct answer -a matrix that is obtained by performing a single elementary row operation on an identity matrix
