Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}{9}^2& -10^2& 11^2\\ 12^2& 13^2& -14^2\\ -15^2& 16^2& 17^2\end{array}\right]$, then the value of $A^{T}BA$ is:

We are given a column vector $A=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and a $3\times 3$ matrix $B$. Our objective is to evaluate the quadratic form represented by the expression $A^{T}BA$.

The matrix $B$ is defined by the squares of integers as follows:

Let us define our components. We have $A=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and its transpose, $A^T=\left[\begin{array}{ccc}1& 1& 1\end{array}\right]$.

When we compute $A^{T}BA$, we are performing the product of a $1\times 3$ row vector, a $3\times 3$ matrix, and a $3\times 1$ column vector. The beauty of this operation is that multiplying by a vector of ones acts as a summation operator.

Specifically, $A^{T}B$ results in a row vector where each element is the sum of the corresponding column of $B$. Multiplying this result by $A$ then sums those column totals together. Thus, $A^{T}BA$ is equivalent to the sum of all elements in matrix $B$.

Let us calculate the sum of each column of $B$:

Column 1: $9^2+12^2-15^2=81+144-225=225-225=0$.

Column 2: $-10^2+13^2+16^2=-100+169+256=69+256=325$.

Column 3: $11^2-14^2+17^2=121-196+289=-75+289=214$.

To find the final value of the expression, we sum the results obtained from the columns:

The value of the expression $A^{T}BA$ is 539.

Always remember that for any matrix $B$, if $A$ is a vector of ones, $A^{T}BA$ is simply the sum of all elements in $B$. Recognizing this property transforms complex matrix multiplication into a simple arithmetic task.