MAT 343 Applied Linear Algebra July 17, 2008
Sample solutions for test 1
1. undefined: a., b. inner dimensions don’t match, e. not square. Matrices: c., d. 3×3, f. 2 ×2.
2.a.
−110
1−2 1
0 1 1
x1
x2
x3
=
3
−5
8
b.
-1 1 0 3
1−2 1 −5
0 1 1 8
R20=R2+R1
−→
-1 1 0 3
0 -1 1 −2
0 1 1 8
R30=R3+R2
−→
-1 1 0 3
0 -1 1 −2
0 0 2 6
.
2z= 6 yields z= 3.
Back-substitution: −y+z=−2 yields y=−(−3−2) = 5, and
−x+y= 3 yields x=−(−5 + 3) = 2.
c.E1=
100
110
001
.d. E2=
100
010
011
. E123 =E2E1=
100
110
111
.
e. L=E−1
1E−1
2=
1 0 0
−110
0 0 1
100
010
0−1 1
=
1 0 0
−110
0−1 1
. U =
−110
0−1 1
0 0 2
(from b.).
3. a. Define CA =A−C= (6,0,−12) and CB =B−C= (0,4,−12).
Then 122=CA ·C B =kCAkkCBkcos γ= 6√5·4√10 ·cos γyields γ≈31.948◦.
b. Calculate the normal vector N= (2,3,1) as a nonzero solution of MN = 0 where Mis the matrix
with rows CA and CB (alternatively, calculate the cross product of CA and CB ).
The desired distance is the length of the projection of Aonto N, i.e. d=A·N/kNk ≈ 2.1381.
c. Write AB =B−A= (−6,4,0) and AC =−CA. The closest point PBis the projection of AB
onto AC, translated by A.PB=A+AB·AC
AC·AC AC =12
5(2,0,1) = (4.8,0,2.4).
4.•Ais nonsingular (different name). •There exists Bsuch that AB =BA =I(definition).
•Ahas npivots. •Uhas no zero row (†).•Uis invertible. •rref(A) has no zero row. •rref(A) = I.
•The only solution of Ax = 0 is x= 0. •For every b∈IRn,Ax =bhas a unique solution.
•ATis invertible. •AThas npivots. •det A6= 0.
And many more to come in terms of e.g. rank, linear independence, spanning, eigenvalues . . .
(†)Uis the upper triangular matrix of the LU-decomposition of A.
5. F T F T T F F T. (i.) A=B=I. (ii.) C=BA−1satisfies C(AB−1) = (AB−1)C=I.
(iii.) A=(1 0)=BT. (iv.) Q2(Q2)T=Q(QQT)QT=QIQT=Iand (Q2)TQ2=QT(QTQ)Q=QTI Q =I.
(v.) Pairwise orthogonal columns means QTQ=I, i.e. QTis a left inverse of Q, and, since Qis square,
also a right inverse. But QQT=Imeans the rows of Qare pairwise orthogonal.
(vi.) A=(1 0), B= (0 2)T,C= (0 3)T. (vii.) 1 1
0 1 !−1
= 1−1
0 1 !is not lower triangular.
(viii.) The only nonzero entry of L2is in its bottom left corner, but the third column of Lis zero.
Hence all columns of L3=L·L2are zero.
♣