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We are given a matrix problem with the following elements:

- A square matrix \( T \) of order 3 is upper triangular, where \( t_{ij} = 2i^2 - j \) for \( i \leq j \).
- Matrix \( A = [-1 \ 1 \ 1] \) is a \( 1 \times 3 \) matrix.
- We are tasked with finding the value of \( A \cdot T \cdot A^t \), where \( A^t \) is the transpose of \( A \), resulting in a \( 1 \times 1 \) matrix.

Let's break this down step-by-step:

### Step 1: Construct the Matrix \( T \)
We will first determine the elements of matrix \( T \) using the formula \( t_{ij} = 2i^2 - j \) for \( i \leq j \), as it's upper triangular. The matrix \( T \) is of order 3, so it will be a \( 3 \times 3 \) matrix:

\[
T = \begin{bmatrix}
t_{11} & t_{12} & t_{13} \\
0      & t_{22} & t_{23} \\
0      & 0      & t_{33}
\end{bmatrix}
\]

We now compute each element:
- \( t_{11} = 2(1)^2 - 1 = 2 - 1 = 1 \)
- \( t_{12} = 2(1)^2 - 2 = 2 - 2 = 0 \)
- \( t_{13} = 2(1)^2 - 3 = 2 - 3 = -1 \)
- \( t_{22} = 2(2)^2 - 2 = 8 - 2 = 6 \)
- \( t_{23} = 2(2)^2 - 3 = 8 - 3 = 5 \)
- \( t_{33} = 2(3)^2 - 3 = 18 - 3 = 15 \)

Thus, the matrix \( T \) is:

\[
T = \begin{bmatrix}
1 & 0 & -1 \\
0 & 6 & 5 \\
0 & 0 & 15
\end{bmatrix}
\]

### Step 2: Calculate \( A \cdot T \)
Now we compute the product \( A \cdot T \), where \( A = [-1 \ 1 \ 1] \).

\[
A \cdot T = [-1 \ 1 \ 1] \cdot \begin{bmatrix}
1 & 0 & -1 \\
0 & 6 & 5 \\
0 & 0 & 15
\end{bmatrix}
\]

Perform the multiplication:

\[
A \cdot T = [-1 \cdot 1 + 1 \cdot 0 + 1 \cdot 0, \ -1 \cdot 0 + 1 \cdot 6 + 1 \cdot 0, \ -1 \cdot (-1) + 1 \cdot 5 + 1 \cdot 15]
\]
\[
= [-1, 6, 1 + 5 + 15] = [-1, 6, 21]
\]

### Step 3: Compute \( A \cdot T \cdot A^t \)
Next, calculate the product \( A \cdot T \cdot A^t \). We already have \( A \cdot T = [-1, 6, 21] \), and \( A^t = \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \).

Now, multiply \( [-1, 6, 21] \cdot \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \):

\[
[-1, 6, 21] \cdot \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} = (-1 \cdot -1) + (6 \cdot 1) + (21 \cdot 1) = 1 + 6 + 21 = 28
\]

### Final Answer:
The result of \( A \cdot T \cdot A^t \) is 28, so the correct option is:

**d) 28**

---

Would you like further clarification on any step? 

Here are five related questions to explore:
1. How do we determine whether a matrix is upper triangular?
2. What is the significance of the transpose operation in matrix multiplication?
3. How does matrix multiplication differ from scalar multiplication?
4. What properties of the transpose matrix were used in this problem?
5. How would the result change if the matrix \( A \) were a \( 3 \times 1 \) vector instead?

**Tip:** Always verify matrix dimensions before performing multiplication to ensure compatibility!

Uma matriz quadrada de ordem n é chamada triangular superior se a_ij = 0 para i > j. Os elementos de uma matriz triangular superior T, de ordem 3, onde i ≤ j, são obtidos a partir da lei de formação tij = 2i² − j. Sendo A = [−1 1 1] uma matriz de ordem 1×3 e A^t sua transposta, o produto A⋅T⋅A^t é a matriz 1×1 cujo único elemento vale

Solve the problem of finding the product A·T·A^t, where A is a 1x3 matrix and T is an upper triangular matrix defined by t_ij = 2i² - j for i ≤ j. This detailed explanation covers key concepts of matrix multiplication and transposition.

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