Question:
Let \(T:R^2\to R^2\) be a linear transform. If \(T(1,2)=(2,3)\) and \(T(0,1)=(1,4)\), then \(T(5,6)=\)
Let \(T:R^2\to R^2\) be a linear transform. If \(T(1,2)=(2,3)\) and \(T(0,1)=(1,4)\), then \(T(5,6)=\)
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For a linear transformation, express the required vector as a linear combination of known vectors.
Updated On: Jul 3, 2026
- \((6,-1)\)
- \((-1,6)\)
- \((-6,1)\)
- \((1,-6)\)
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The Correct Option is A
Solution and Explanation
Concept:
Since \(T\) is linear, \[ T(\alpha u+\beta v)=\alpha T(u)+\beta T(v) \] We express \((5,6)\) as a linear combination of \((1,2)\) and \((0,1)\).
Step 1: Express \((5,6)\). Let \[ (5,6)=a(1,2)+b(0,1) \] Then, \[ a(1,2)+b(0,1)=(a,2a+b) \] So, \[ a=5 \] and \[ 2a+b=6 \] Substitute \(a=5\): \[ 10+b=6 \] \[ b=-4 \] Thus, \[ (5,6)=5(1,2)-4(0,1) \]
Step 2: Apply linearity. \[ T(5,6)=5T(1,2)-4T(0,1) \] \[ =5(2,3)-4(1,4) \] \[ =(10,15)-(4,16) \] \[ =(6,-1) \]
Step 3: Final answer. \[ \boxed{(6,-1)} \]
Since \(T\) is linear, \[ T(\alpha u+\beta v)=\alpha T(u)+\beta T(v) \] We express \((5,6)\) as a linear combination of \((1,2)\) and \((0,1)\).
Step 1: Express \((5,6)\). Let \[ (5,6)=a(1,2)+b(0,1) \] Then, \[ a(1,2)+b(0,1)=(a,2a+b) \] So, \[ a=5 \] and \[ 2a+b=6 \] Substitute \(a=5\): \[ 10+b=6 \] \[ b=-4 \] Thus, \[ (5,6)=5(1,2)-4(0,1) \]
Step 2: Apply linearity. \[ T(5,6)=5T(1,2)-4T(0,1) \] \[ =5(2,3)-4(1,4) \] \[ =(10,15)-(4,16) \] \[ =(6,-1) \]
Step 3: Final answer. \[ \boxed{(6,-1)} \]
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