Question

Let $T: \mathbb{R}^3 \to \mathbb{R}^4$ be a linear transformation. If $T(1,1,0) = (2,0,0,0)$, $T(1,0,1) = (2,4,0,0)$, and $T(0,1,1) = (0,0,2,0)$, then $T(1,1,1)$ equals
 

• A. $(1,1,0,0)$
• B. $(0,1,1,1)$
• C. $(2,2,1,0)$
• D. $(0,0,0,0)$

Hint:

For linear transformations, use linear combinations of known images and the property $T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})$.

Answer 1

The Correct Option is C

Step 1: Express $(1,1,1)$ as a linear combination. 
We have vectors: \[ v_1 = (1,1,0), v_2 = (1,0,1), v_3 = (0,1,1). \] We find $a,b,c$ such that \[ a(1,1,0) + b(1,0,1) + c(0,1,1) = (1,1,1). \] This gives system: \[ \begin{cases} a + b = 1, \\ a + c = 1, \\ b + c = 1. \end{cases} \] Solving, $a = b = c = \frac{1}{2}$. 
 

Step 2: Use linearity of $T$. 
\[ T(1,1,1) = \frac{1}{2}[T(1,1,0) + T(1,0,1) + T(0,1,1)]. \] \[ = \frac{1}{2}[(2,0,0,0) + (2,4,0,0) + (0,0,2,0)] = \frac{1}{2}(4,4,2,0) = (2,2,1,0). \]

Step 3: Conclusion. 
\[ \boxed{T(1,1,1) = (2,2,1,0)}. \]