Linear Transformation and Bases In Exercises 25-28, let T : R 3 → R 3 be a linear transformation such that T ( 1 , 0 , 0 ) = ( 2 , 4 , - 1 ) , T ( 0 , 1 , 0 ) = ( 1 , 3 , - 2 ) , and T ( 0 , 0 , 1 ) = ( 0 , - 2 , 2 ) . Find the specified image. T ( − 2 , 4 , − 1 )
Linear Transformation and Bases In Exercises 25-28, let T : R 3 → R 3 be a linear transformation such that T ( 1 , 0 , 0 ) = ( 2 , 4 , - 1 ) , T ( 0 , 1 , 0 ) = ( 1 , 3 , - 2 ) , and T ( 0 , 0 , 1 ) = ( 0 , - 2 , 2 ) . Find the specified image. T ( − 2 , 4 , − 1 )
Solution Summary: The author explains how the vector (-2,4,-1) can be written as a vector.
Linear Transformation and BasesIn Exercises 25-28, let
T
:
R
3
→
R
3
be a linear transformation such that
T
(
1
,
0
,
0
)
=
(
2
,
4
,
-
1
)
,
T
(
0
,
1
,
0
)
=
(
1
,
3
,
-
2
)
, and
T
(
0
,
0
,
1
)
=
(
0
,
-
2
,
2
)
. Find the specified image.
Solve the system of equation for y using Cramer's rule. Hint: The
determinant of the coefficient matrix is -23.
-
5x + y − z = −7
2x-y-2z = 6
3x+2z-7
eric
pez
Xte
in
z=
Therefore, we have
(x, y, z)=(3.0000,
83.6.1 Exercise
Gauss-Seidel iteration with
Start with (x, y, z) = (0, 0, 0). Use the convergent Jacobi i
Tol=10 to solve the following systems:
1.
5x-y+z = 10
2x-8y-z=11
-x+y+4z=3
iteration (x
Assi 2
Assi 3.
4.
x-5y-z=-8
4x-y- z=13
2x - y-6z=-2
4x y + z = 7
4x-8y + z = -21
-2x+ y +5z = 15
4x + y - z=13
2x - y-6z=-2
x-5y- z=-8
realme Shot on realme C30
2025.01.31 22:35
f
Use Pascal's triangle to expand the binomial
(6m+2)^2
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Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY