Question

Let $T:R^{3}\rightarrow R^{3}$ be a linear transformation defined by $T(a,b,c)=(a+b-c, a+b+c, b-c)$. Then the matrix of $T$ with respect to the ordered basis $B = \{(0,1,0), (0,0,1), (1,0,0)\}$ is}

• A. $\left[\begin{matrix}1& 1&-1\\ 1& 1& 1\\ 0& 1&-1\end{matrix}\right]$
• B. $\left[\begin{matrix}1& 1& 0\\ 1& 1& 1\\ 1& 0&-1\end{matrix}\right]$
• C. $\left[\begin{matrix}1&-1& 1
  1& 1& 1
  1&-1& 0\end{matrix}\right]$
• D. $\left[\begin{matrix}1& 1& 1
  1&-1& 0
  1&-1& 1\end{matrix}\right]$

Hint:

Matrix of $T = [ [T(v_1)]_B | [T(v_2)]_B | [T(v_3)]_B ]$. The columns are images of the basis vectors.

Answer 1

The Correct Option is C

Step 1: Concept
To find the matrix, apply $T$ to each basis vector and express the result as a coordinate vector relative to the same basis $B$.

Step 2: Meaning
$B = \{v_1, v_2, v_3\}$ where $v_1=(0,1,0), v_2=(0,0,1), v_3=(1,0,0)$. $T(v_1) = T(0,1,0) = (1, 1, 1) = 1v_1 + 1v_2 + 1v_3$. $T(v_2) = T(0,0,1) = (-1, 1, -1) = -1v_1 + 1v_2 + 1v_3$. $T(v_3) = T(1,0,0) = (1, 1, 0) = 1v_1 + 1v_2 + 0v_3$.

Step 3: Analysis
The coordinate vectors are $(1,1,1)^T, (-1,1,1)^T$, and $(1,1,0)^T$. Note the basis order in the prompt is $(0,1,0), (0,0,1), (1,0,0)$.

Step 4: Conclusion
Placing these as columns, the matrix is $\left[\begin{matrix}1&-1& 1\\ 1& 1& 1\\ 1&-1& 0\end{matrix}\right]$. Final Answer: (C)