Explanation Given: The linear transformation, T ( x 2 − 1 ) = x 2 + x − 3 , T ( 2 x ) = 4 x , T ( 3 x + 2 ) = 2 ( x + 3 ) . Approach: Let V and W be vector spaces. A mapping T : V → W is called a linear transformation form V to W if it satisfies the following properties: 1. T ( u + v ) = T ( u ) + T ( v ) for all u , v ∈ V . 2. T ( c v ) = c T ( v ) for all v ∈ V and all scalars c . It refers to these properties as the linearity properties. The vector space V is called the domain of T , while the vector space W is called the co-domain of T . Calculation: Calculate the transformation T ( 1 ) . Since from the transformation, 1 is the combination of ( x 2 − 1 ) , ( 2 x ) and ( x + 3 ) then, 1 = a ( x 2 − 1 ) + b ( 2 x ) + c ( x + 3 ) ... ( 1 ) Here, a , b and c are real number. The above equation can be written as, 1 = ( a ) x 2 + ( 2 b + 3 c ) x + ( − a + 2 c ) The coefficient of the above equation is written as, a = 0 ... ( 2 ) 2 b + 3 c = 0 ... ( 3 ) − a + 2 c = 1 ... ( 4 ) Solve equation ( 2 ) , equation ( 3 ) and equation ( 4 ) to get the value of a , b and c . The value of a = 0 , b = − 3 4 and c = 1 2 . Substitute 0 for a , − 3 4 for b and 1 2 for c in the equation ( 1 ) . 1 = 0 ( x 2 − 1 ) − 3 4 ( 2 x ) + 1 2 ( x + 3 ) Apply transformation on both the side of the above equation. T ( 1 ) = T [ 0 ( x 2 − 1 ) − 3 4 ( 2 x ) + 1 2 ( x + 3 ) ] = − 3 4 T ( 2 x ) + 1 2 T ( x + 3 ) Substitute 4 x for T ( 2 x ) and 2 ( x + 3 ) for T ( 3 x + 2 ) in the above equation. T ( 1 ) = − 3 4 ( 4 x ) + 1 2 [ 2 ( x + 3 ) ] = − 3 x + x + 3 = − 2 x + 3 So, the linear transformation T ( 1 ) = − 2 x + 3 . Calculate the transformation T ( x ) . Since from the transformation, x is the combination of ( x 2 − 1 ) , ( 2 x ) and ( x + 3 ) then, x = a ( x 2 − 1 ) + b ( 2 x ) + c ( x + 3 ) ... ( 5 ) Here, a , b and c are real number. The above equation can be written as, x = ( a ) x 2 + ( 2 b + 3 c ) x + ( − a + 2 c ) The coefficient of the above equation is written as, a = 0 ... ( 6 ) 2 b + 3 c = 1 ... ( 7 ) − a + 2 c = 0 ... ( 8 ) Solve equation ( 6 ) , equation ( 7 ) and equation ( 8 ) to get the value of a , b and c . The value of a = 0 , b = 1 2 and c = 0 . Substitute 0 for a , 1 2 for b and 0 for c in the equation ( 5 ) . x = 0 ( x 2 − 1 ) + 1 2 ( 2 x ) + 0 ( x + 3 ) Apply transformation on both the side of the above equation. T ( x ) = T [ 0 ( x 2 − 1 ) + 1 2 ( 2 x ) + 0 ( x + 3 ) ] = 1 2 T ( 2 x ) Substitute 4 x for T ( 2 x ) in the above equation

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 6.1, Problem 33P

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The values of linear transformation T(1), T(x), T(x2), and show that T(ax2+bx+c)=ax2−(a−2b+2c)x+3c,

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Chapter 6.1, Problem 32P

Chapter 6.1, Problem 34P

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Chapter 6 Solutions

Differential Equations and Linear Algebra (4th Edition)

Ch. 6.1 - True-False Review For Questions a-f, decide if the...Ch. 6.1 - True-False Review For Questions a-f, decide if the...Ch. 6.1 - True-False Review For Questions a-f, decide if the...Ch. 6.1 - True-False Review For Questions a-f, decide if the...Ch. 6.1 - True-False Review For Questions a-f, decide if the...Ch. 6.1 - True-False Review For Questions a-f, decide if the...Ch. 6.1 - For problem 1-8, verify directly from Definition...Ch. 6.1 - For problems 1-8, verify directly from Definition...Ch. 6.1 - For problem 1-8, verify directly from Definition...Ch. 6.1 - For problem 1-8, verify directly from Definition...

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The values of linear transformation T ( 1 ) , T ( x ) , T ( x 2 ) , and show that T ( a x 2 + b x + c ) = a x 2 − ( a − 2 b + 2 c ) x + 3 c ,

The values of linear transformation T(1), T(x), T(x2), and show that T(ax2+bx+c)=ax2−(a−2b+2c)x+3c,

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Chapter 6 Solutions