Question:
If \(T_1:R^3\to R^2\) and \(T_2:R^2\to R^2\) are two linear transformations defined by \(T_1(x,y,z)=(3x,4y-z)\) and \(T_2(x,y)=(-x,y)\), then \(T_2T_1\) is
If \(T_1:R^3\to R^2\) and \(T_2:R^2\to R^2\) are two linear transformations defined by \(T_1(x,y,z)=(3x,4y-z)\) and \(T_2(x,y)=(-x,y)\), then \(T_2T_1\) is
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For \(T_2T_1\), always apply \(T_1\) first and then \(T_2\).
Updated On: Jul 3, 2026
- \((-3x,4y+z)\)
- \((-3x,4y-z)\)
- \((3x,4y+z)\)
- not defined
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The Correct Option is B
Solution and Explanation
Concept:
The composition \(T_2T_1\) means \[ T_2T_1=T_2\circ T_1 \] So we first apply \(T_1\), then apply \(T_2\).
Step 1: Apply \(T_1\). \[ T_1(x,y,z)=(3x,4y-z) \] This gives an ordered pair in \(R^2\).
Step 2: Apply \(T_2\). The transformation \(T_2\) is \[ T_2(u,v)=(-u,v) \] Here, \[ u=3x,\qquad v=4y-z \] Therefore, \[ T_2(3x,4y-z)=(-3x,4y-z) \]
Step 3: Final answer. \[ \boxed{(-3x,4y-z)} \]
The composition \(T_2T_1\) means \[ T_2T_1=T_2\circ T_1 \] So we first apply \(T_1\), then apply \(T_2\).
Step 1: Apply \(T_1\). \[ T_1(x,y,z)=(3x,4y-z) \] This gives an ordered pair in \(R^2\).
Step 2: Apply \(T_2\). The transformation \(T_2\) is \[ T_2(u,v)=(-u,v) \] Here, \[ u=3x,\qquad v=4y-z \] Therefore, \[ T_2(3x,4y-z)=(-3x,4y-z) \]
Step 3: Final answer. \[ \boxed{(-3x,4y-z)} \]
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