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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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306 Chapter 8 Matrix algebra

Thus we have

2x+2y =0

−x +z=0

3x+2y −z=0

Wenotethatthethirdequationcanbederivedfromthefirsttwoequations,bysubtracting

thesecondequationfromthefirst.Ifyoucannotspotthistheequationsshouldbesolved

by Gaussian elimination. Ineffect we have only two equations:

2x+2y =0

−x +z =0

Solving these givesx =t,y = −t,z =t. The eigenvector is

⎛ ⎞

1

X =t ⎝−1⎠

1

λ = 1

We have

⎡⎛

1 2 0

⎛ ⎞⎤

⎛ ⎞

100 x

⎛ ⎞

0

⎣⎝−1−1 1⎠ − ⎝010⎠⎦

⎝y⎠ = ⎝0⎠

3 2−2 001 z 0

Thus we have

2y =0

−x−2y+ z=0

3x +2y−3z=0

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

0 2 0 x 0

⎝−1−2 1⎠

⎝y⎠ = ⎝0⎠

3 2−3 z 0

From the first equation,y = 0;puttingy = 0 into the other equations yields

−x+z =0

3x−3z =0

Here the second equation can be derived from the first by multiplying the first by −3.

Solving, wehavex =t,z =t. So the eigenvector is

⎛ ⎞

1

X =t ⎝0⎠

1

EXERCISES8.11.3

1 Calculate the eigenvectors ofthe matrices given in

Question 1 ofExercises8.11.2.

2 Calculate the eigenvectors ofthe matrices given in

Question2ofExercises8.11.2.

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