Exercise 6 (1 point). If T: mathbb{R}^{4} rightarrow mathbb{R}^{3} is a linear transformation such that Tleft(begin{array}{l}1 1 1 1end{array}right)=left(begin{array}{c}5 1 -3end{array}right) quad and Tleft(begin{array}{c}-1 1 0 2end{array}right)=left(begin{array}{l}2 0 1end{array}right) operatorname{find} Tleft(begin{array}{c}5 -1 2 -4end{array}right)

Question

Exercise 6 (1 point). If T: mathbb{R}^{4} rightarrow mathbb{R}^{3} is a linear transformation such that Tleft(begin{array}{l}1  1  1  1end{array}right)=left(begin{array}{c}5  1  -3end{array}right) quad and Tleft(begin{array}{c}-1  1  0  2end{array}right)=left(begin{array}{l}2  0  1end{array}right) operatorname{find} Tleft(begin{array}{c}5  -1  2  -4end{array}right)

Accepted Answer

Step:1 ：First, we try to express (begin{pmatrix} 5  -1  2  -4 end{pmatrix}) as a linear combination of (begin{pmatrix} 1  1  1  1 end{pmatrix}) and (begin{pmatrix} -1  1  0  2 end{pmatrix}):Step:2 ：(begin{pmatrix} 5  -1  2  -4 end{pmatrix} = a begin{pmatrix} 1  1  1  1 end{pmatrix} + b begin{pmatrix} -1  1  0  2 end{pmatrix} = begin{pmatrix} a - b  a + b  a  a + 2b end{pmatrix})Step:3 ：Solving the system of equations (a - b = 5), (a + b = -1), (a = 2), and (a + 2b = -4), we find that (a = 2) and (b = -3).Step:4 ：Thus, (begin{pmatrix} 5  -1  2  -4 end{pmatrix} = 2 begin{pmatrix} 1  1  1  1 end{pmatrix} - 3 begin{pmatrix} -1  1  0  2 end{pmatrix}).Step:5 ：By the linearity of the transformation (T), we have (T begin{pmatrix} 5  -1  2  -4 end{pmatrix} = 2 T begin{pmatrix} 1  1  1  1 end{pmatrix} - 3 T begin{pmatrix} -1  1  0  2 end{pmatrix}).Step:6 ：Substituting the given values of (Tbegin{pmatrix} 1  1  1  1 end{pmatrix}) and (Tbegin{pmatrix} -1  1  0  2 end{pmatrix}), we get (T begin{pmatrix} 5  -1  2  -4 end{pmatrix} = 2 begin{pmatrix} 5  1  -3 end{pmatrix} - 3 begin{pmatrix} 2  0  1 end{pmatrix}).Step:7 ：Calculating the result, we find (T begin{pmatrix} 5  -1  2  -4 end{pmatrix} = boxed{begin{pmatrix} 4  2  -7 end{pmatrix}}).

Answer

Step: 1 ：First, we try to express (begin{pmatrix} 5  -1  2  -4 end{pmatrix}) as a linear combination of (begin{pmatrix} 1  1  1  1 end{pmatrix}) and (begin{pmatrix} -1  1  0  2 end{pmatrix}):Step: 2 ：(begin{pmatrix} 5  -1  2  -4 end{pmatrix} = a begin{pmatrix} 1  1  1  1 end{pmatrix} + b begin{pmatrix} -1  1  0  2 end{pmatrix} = begin{pmatrix} a - b  a + b  a  a + 2b end{pmatrix})Step: 3 ：Solving the system of equations (a - b = 5), (a + b = -1), (a = 2), and (a + 2b = -4), we find that (a = 2) and (b = -3).Step: 4 ：Thus, (begin{pmatrix} 5  -1  2  -4 end{pmatrix} = 2 begin{pmatrix} 1  1  1  1 end{pmatrix} - 3 begin{pmatrix} -1  1  0  2 end{pmatrix}).Step: 5 ：By the linearity of the transformation (T), we have (T begin{pmatrix} 5  -1  2  -4 end{pmatrix} = 2 T begin{pmatrix} 1  1  1  1 end{pmatrix} - 3 T begin{pmatrix} -1  1  0  2 end{pmatrix}).Step: 6 ：Substituting the given values of (Tbegin{pmatrix} 1  1  1  1 end{pmatrix}) and (Tbegin{pmatrix} -1  1  0  2 end{pmatrix}), we get (T begin{pmatrix} 5  -1  2  -4 end{pmatrix} = 2 begin{pmatrix} 5  1  -3 end{pmatrix} - 3 begin{pmatrix} 2  0  1 end{pmatrix}).Step: 7 ：Calculating the result, we find (T begin{pmatrix} 5  -1  2  -4 end{pmatrix} = boxed{begin{pmatrix} 4  2  -7 end{pmatrix}}).

Answer

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