Find the lineal transformations given 2 basis and the associated matrix
$begingroup$
Translation:
Being T: R3 to R2 a lineal transformation. Basis 1: ..... and Basis 2:...... ,Basis of R3 and R2 respectively. Consider:
A (associated matrix)= T B1 to B2=.....
Find:
a)T(1,1,0)B2
b) T(3,2,1)
linear-algebra
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
$endgroup$
add a comment |
$begingroup$
Translation:
Being T: R3 to R2 a lineal transformation. Basis 1: ..... and Basis 2:...... ,Basis of R3 and R2 respectively. Consider:
A (associated matrix)= T B1 to B2=.....
Find:
a)T(1,1,0)B2
b) T(3,2,1)
linear-algebra
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
$endgroup$
add a comment |
$begingroup$
Translation:
Being T: R3 to R2 a lineal transformation. Basis 1: ..... and Basis 2:...... ,Basis of R3 and R2 respectively. Consider:
A (associated matrix)= T B1 to B2=.....
Find:
a)T(1,1,0)B2
b) T(3,2,1)
linear-algebra
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
$endgroup$
Translation:
Being T: R3 to R2 a lineal transformation. Basis 1: ..... and Basis 2:...... ,Basis of R3 and R2 respectively. Consider:
A (associated matrix)= T B1 to B2=.....
Find:
a)T(1,1,0)B2
b) T(3,2,1)
linear-algebra
linear-algebra
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
asked Dec 16 '18 at 18:52
Nelson AguileraNelson Aguilera
153
153
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Write $(1,1,0)$ in the form of $B_1$ basis which is equal to $(0,1,0)_{B_1}$ . Now multiply it with $A$ to get
$$|T(1,1,0)|_{B_2} = Acdot (0,1,0)^T = (0,1)_{B_2}$$
For part b), write $(3,2,1)$ in the form of $B_1$ basis which is equal to $(1,1,1)_{B_1}$. Now multiply this with $A$ to get
$$T(3,2,1) = Acdot (1,1,1)^T=(0,2)_{B_2} = (2,-2)$$
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
$endgroup$
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write $(1,1,0)$ in the form of $B_1$ basis which is equal to $(0,1,0)_{B_1}$ . Now multiply it with $A$ to get
$$|T(1,1,0)|_{B_2} = Acdot (0,1,0)^T = (0,1)_{B_2}$$
For part b), write $(3,2,1)$ in the form of $B_1$ basis which is equal to $(1,1,1)_{B_1}$. Now multiply this with $A$ to get
$$T(3,2,1) = Acdot (1,1,1)^T=(0,2)_{B_2} = (2,-2)$$
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
$endgroup$
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
add a comment |
$begingroup$
Write $(1,1,0)$ in the form of $B_1$ basis which is equal to $(0,1,0)_{B_1}$ . Now multiply it with $A$ to get
$$|T(1,1,0)|_{B_2} = Acdot (0,1,0)^T = (0,1)_{B_2}$$
For part b), write $(3,2,1)$ in the form of $B_1$ basis which is equal to $(1,1,1)_{B_1}$. Now multiply this with $A$ to get
$$T(3,2,1) = Acdot (1,1,1)^T=(0,2)_{B_2} = (2,-2)$$
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
$endgroup$
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
add a comment |
$begingroup$
Write $(1,1,0)$ in the form of $B_1$ basis which is equal to $(0,1,0)_{B_1}$ . Now multiply it with $A$ to get
$$|T(1,1,0)|_{B_2} = Acdot (0,1,0)^T = (0,1)_{B_2}$$
For part b), write $(3,2,1)$ in the form of $B_1$ basis which is equal to $(1,1,1)_{B_1}$. Now multiply this with $A$ to get
$$T(3,2,1) = Acdot (1,1,1)^T=(0,2)_{B_2} = (2,-2)$$
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
$endgroup$
Write $(1,1,0)$ in the form of $B_1$ basis which is equal to $(0,1,0)_{B_1}$ . Now multiply it with $A$ to get
$$|T(1,1,0)|_{B_2} = Acdot (0,1,0)^T = (0,1)_{B_2}$$
For part b), write $(3,2,1)$ in the form of $B_1$ basis which is equal to $(1,1,1)_{B_1}$. Now multiply this with $A$ to get
$$T(3,2,1) = Acdot (1,1,1)^T=(0,2)_{B_2} = (2,-2)$$
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
answered Dec 16 '18 at 19:25
Sauhard SharmaSauhard Sharma
953318
953318
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
add a comment |
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
$begingroup$
yeah thanks, both of your answers are right.
$endgroup$
– Nelson Aguilera
Dec 16 '18 at 19:56
add a comment |
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