Question

Consider the real vector space \( \mathbb{R}^3 \). Let \( T : \mathbb{R}^3 \to \mathbb{R} \) be a linear transformation such that \[ T(1, 1, 1) = 0, \quad T(1, -1, 1) = 0, \quad T(0, 0, 1) = 16. \] Then, the value of \( T \left( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \right) \) is equal to ............... (rounded off to two decimal places).

Hint:

For linear transformations, express the vector as a linear combination of given vectors and apply the transformation to each term.

Answer 1

Step 1: Express \( T(x, y, z) \) as a linear combination of the given vectors.
We are given three vectors \( (1, 1, 1) \), \( (1, -1, 1) \), and \( (0, 0, 1) \). Since \( T \) is a linear transformation, we can express any vector \( (x, y, z) \) as a linear combination of these three vectors. Let's assume: \[ \left( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \right) = \alpha (1, 1, 1) + \beta (1, -1, 1) + \gamma (0, 0, 1). \] Now, we solve for \( \alpha \), \( \beta \), and \( \gamma \). Step 2: Set up the system of equations.
We equate the components: \[ \frac{1}{2} = \alpha + \beta, \quad \frac{2}{3} = \alpha - \beta, \quad \frac{3}{4} = \alpha + \gamma. \] Step 3: Solve the system.
From the second equation, \( \alpha - \beta = \frac{2}{3} \), solve for \( \beta \): \[ \beta = \alpha - \frac{2}{3}. \] Substitute this into the first equation: \[ \frac{1}{2} = \alpha + \left( \alpha - \frac{2}{3} \right) = 2\alpha - \frac{2}{3}. \] Solving for \( \alpha \): \[ 2\alpha = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}, \quad \alpha = \frac{7}{12}. \] Substitute \( \alpha = \frac{7}{12} \) into \( \beta = \alpha - \frac{2}{3} \): \[ \beta = \frac{7}{12} - \frac{8}{12} = -\frac{1}{12}. \] Now, substitute \( \alpha = \frac{7}{12} \) into the third equation: \[ \frac{3}{4} = \frac{7}{12} + \gamma, \quad \gamma = \frac{3}{4} - \frac{7}{12} = \frac{5}{12}. \] Step 4: Apply the linear transformation.
Now that we have \( \alpha = \frac{7}{12} \), \( \beta = -\frac{1}{12} \), and \( \gamma = \frac{5}{12} \), we can compute \( T \left( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \right) \): \[ T \left( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \right) = \alpha T(1, 1, 1) + \beta T(1, -1, 1) + \gamma T(0, 0, 1). \] Substitute the given values for \( T(1, 1, 1) = 0 \), \( T(1, -1, 1) = 0 \), and \( T(0, 0, 1) = 16 \): \[ T \left( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \right) = \frac{7}{12} \cdot 0 + \left( -\frac{1}{12} \right) \cdot 0 + \frac{5}{12} \cdot 16 = \frac{5}{12} \cdot 16 = \frac{80}{12} = \frac{20}{3} \approx 6.67. \] Final Answer: \[ \boxed{6.67}. \]