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Problem A and pg.100 #3 are graded.
Problem A: Find the image of the square whose corners lie at the points (0,0),(1,0),(0,1),(1,1) in the plane under the linear transformation
T:R2×1→R2×1T(→x)=A→x,
where A is the matrix A=[1501].
Solution: We interpret the four corners of the square as vectors and plug them into the transformation T: first plug in [00] and get
T([00])=[00].
Now plug in [10] to get
T([10])=[10].
Now plug in [01] to get
T([01])=[51]
Finally, plug in [11] to get
T([11])=[61].
Thus you get the following picture:
#3,pg.100: Let A=[2−53−2]. Compute 3I2−A and (3I2)A.
Solution: First we compute
3I2−A=[3003]−[2−53−2]=[15−35].
Now compute
(3I2)A=[3003][2−53−2]=[6−159−6]
