Linear Algebra with Applications (3rd Edition)

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Chapter 3 SSM: Linear Algebra Chapter 3 3.1 1. Find all ~x such that A~x = ~ 0: " 1 2 . . . 0 3 4 . . . 0 # -→ " 1 0 . . . 0 0 1 . . . 0 # , so that x 1 = x 2 = 0. ker( A ) = { ~ 0 } . 3. Find all ~x such that A~x = ~ 0; note that all ~x in R 2 satisfy the equation, so that ker( A ) = R 2 = span( ~ e 1 ,~ e 2 ). 5. Find all ~x such that A~x = ~ 0. 1 1 1 . . . 0 1 2 3 . . . 0 1 3 5 . . . 0 -→ 1 0 - 1 . . . 0 0 1 2 . . . 0 0 0 0 . . . 0 ; x 1 = x 3 x 2 = - 2 x 3 ; x 1 x 2 x 3 = t - 2 t t ker( A ) = span 1 - 2 1 . 7. Find all ~x such that A~x = ~ 0. Since rref( A ) = I 3 we have ker( A ) = { ~ 0 } . 9. Find all ~x such that A~x = ~ 0. Solving this system yields ker( A ) = { ~ 0 } . 11. Solving the system A~x = ~ 0 we find that ker( A ) = span - 2 3 1 0 . 13. Solving the system A~x = ~ 0 we find that ker( A ) = span - 2 1 0 0 0 0 , - 3 0 - 2 - 1 1 0 , 0 0 0 0 0 1 . 15. By Fact 3.1.3, the image of A is the span of the columns of A : im( A ) = span 1 1 , 1 2 , 1 3 , 1 4 . 64
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SSM: Linear Algebra Section 3.1 Since any two of these vectors span all of R 2 already, we can write im( A ) = span 1 1 , 1 2 . 17. By Fact 3.1.3, im( A ) = span 1 3 , 2 4 = R 2 (the whole plane). 19. Since the four column vectors of A are parallel, we have im( A ) = span 1 - 2 , a line in R 2 . 21. By Fact 3.1.3, im( A ) = span 4 1 5 , 7 9 6 , 3 2 8 . It is hard to tell by inspection whether this span is a plane in R 3 or all of R 3 ; we need to find out whether the third column vector is a linear combination of the first two. 4 7 . . . 3 1 9 . . . 2 5 6 . . . 8 -→ 1 0 . . . 0 0 1 . . . 0 0 0 . . . 1 This shows that the third column vector is not contained in the span of the first two, so that im( A ) = R 3 . 23. im( T ) = R 2 and ker( T ) = { ~ 0 } , since T is invertible (see Summary 3.1.8). 25. im( T ) = R 2 and ker( T ) = { ~ 0 } , since T is invertible (see Summary 3.1.8). 27. Let f ( x ) = x 3 - x = x ( x 2 - 1) = x ( x - 1)( x + 1). Then im( f ) = R , since lim x →∞ f ( x ) = and lim x →-∞ f ( x ) = -∞ but the function fails to be invertible since the equation f ( x ) = 0 has three solutions, x = 0, 1, and - 1. 29. Use spherical coordinates (see any good text on multivariable calculus): f φ θ = sin( φ ) cos( θ ) sin( φ ) sin( θ ) cos( φ ) 65
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Chapter 3 SSM: Linear Algebra 31. The plane x + 3 y + 2 z = 0 is spanned by the two vectors - 2 0 1 and - 3 1 0 , for example. Therefore, A = - 2 - 3 0 1 1 0 does the job. There are many other correct answers. 33. The plane is the kernel of the linear transformation T x y z = x + 2 y + 3 z from R 3 to R . 35. kernel( T ) = { ~x : T ( ~x ) = ~v · ~x = 0 } = the plane with normal vector ~v . im( T ) = R , since for every real number k there is a vector ~x such that T ( ~x ) = k , for example, ~x = k ~v · ~v ~v . 37. A = 0 1 0 0 0 1 0 0 0 , A 2 = 0 0 1 0 0 0 0 0 0 , A 3 = 0 0 0 0 0 0 0 0 0 , so that ker( A ) = span( ~ e 1 ), ker( A 2 ) = span( ~ e 1 ,~ e 2 ), ker( A 3 ) = R 3 , and im( A ) = span( ~ e 1 ,~ e 2 ), im( A 2 ) = span( ~ e 1 ), im( A 3 ) = { ~ 0 } .
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