If `A=[(1),(1),(1)] , B =[(9^2,10^2,11^2),(12^2,-13^2,14^2),(15^2,16^2,-17^2)]` , then `A'BA` is equal to

Options

A. `[665]`
B. `[765]`
C. `[165]`
D. `[365]`

Answer

To solve the problem, we need to compute \( A'BA \), where \( A \) and \( B \) are given matrices.

Given:
\[ A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
\[ B = \begin{pmatrix} 9^2 & 10^2 & 11^2 \\ 12^2 & -13^2 & 14^2 \\ 15^2 & 16^2 & -17^2 \end{pmatrix} \]

### Step 1: Calculate \( A' \) (the transpose of \( A \))
The transpose of matrix \( A \) is:
\[ A' = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A'B \)
Now, we need to multiply \( A' \) with \( B \):
\[ A'B = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 9^2 & 10^2 & 11^2 \\ 12^2 & -13^2 & 14^2 \\ 15^2 & 16^2 & -17^2 \end{pmatrix} \]

Calculating each element:
- First element:
\[ 1 \cdot 9^2 + 1 \cdot 12^2 + 1 \cdot 15^2 = 81 + 144 + 225 = 450 \]
- Second element:
\[ 1 \cdot 10^2 + 1 \cdot (-13^2) + 1 \cdot 16^2 = 100 - 169 + 256 = 187 \]
- Third element:
\[ 1 \cdot 11^2 + 1 \cdot 14^2 + 1 \cdot (-17^2) = 121 + 196 - 289 = 28 \]

Thus, we have:
\[ A'B = \begin{pmatrix} 450 & 187 & 28 \end{pmatrix} \]

### Step 3: Calculate \( A'BA \)
Now we need to multiply \( A'B \) with \( A \):
\[ A'BA = \begin{pmatrix} 450 & 187 & 28 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]

Calculating this gives:
\[ 450 \cdot 1 + 187 \cdot 1 + 28 \cdot 1 = 450 + 187 + 28 = 665 \]

### Final Result
Thus, the value of \( A'BA \) is:
\[ \boxed{665} \]