1. Find the inverse of the following matrices, if they exist.
A=[2111]
A−1=[1−1−12]
B=[32−1501]
B has no inverse.
C=122011−223
C−1=−12−22−11−22−1
D=[−316−2]
D has no inverse.
E=1−112−100−12
E−1=2−314−222−11
F=10003100−211001−41
F−1=1000−3100−5−110−1−541
2. Find the solution to the following linear system using inverses:
x1−2x3=5
2x1+x2+2x3=1
2x1+x2+3x3=−2
x=1−1−2
T:R3→R3 and use it to find a vector x such that T(x)=y where
Tx1x2x3=x1+3x2+x3x2+x32x1−3x3
T−1y1y2y3=63y1+3y2+y3y2+y32y1−3y2−y3
a. y=0−22
x=−1−22
b. y=135
x=−1/38−7/2
4. Use a partitioning of A to find the inverse.
A=200053042
A−1=1/20002/5−3/50−4/55/5
A2=PD2P−1
A3=PD3P−1
An=PDnP−1