Let A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} and B = \begin{pmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{pmatrix}, then the value of A'BA is:
Step-by-Step Solution
Key Concept: Compute the product A'BA where A' is the transpose of A. A'BA = [1 1 1] * B * [1 1 1]^T. This is equivalent to the sum of all elements of matrix B.
A'BA = [1 1 1] * [[9^2, -10^2, 11^2], [12^2, 13^2, -14^2], [-15^2, 16^2, 17^2]] * [1 1 1]^T. Let R be the row vector [1 1 1] * B = [9^2+12^2-15^2, -10^2+13^2+16^2, 11^2-14^2+17^2] = [81+144-225, -100+169+256, 121-196+289] = [0, 325, 214]. Then R * [1 1 1]^T = 0 + 325 + 214 = 539.
Correct Answer: 4