Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

(a)

The given matrix is\(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\), and the given vector is \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\).

Find \({{\bf{x}}^T}A{\bf{x}}\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{5{x_1} + \frac{1}{3}{x_2}}\\{\frac{1}{3}{x_1} + {x_2}}\end{aligned}} \right)\\ &= {x_1}\left( {5{x_1} + \frac{1}{3}{x_2}} \right) + {x_2}\left( {\frac{1}{3}{x_1} + {x_2}} \right)\\ &= 5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2{\rm{ }}\left( 1 \right)\end{aligned}\)

Hence, the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\).

(b)

As the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) when \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\).

So, for \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\), simplify \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) by substituting \({x_1} = 6,{x_2} = 1\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 5{\left( 6 \right)^2} + \left( {\frac{2}{3}} \right)\left( 6 \right)\left( 1 \right) + {\left( 1 \right)^2}\\ &= 185\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 185.

(c)

For \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\), simplify \(5x_1^2 + \left( {\frac{2}{3}} \right){x_1}{x_2} + x_2^2\) by substituting \({x_1} = 1,{x_2} = 3\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 5{\left( 1 \right)^2} + \left( {\frac{2}{3}} \right)\left( 1 \right)\left( 3 \right) + {\left( 3 \right)^2}\\ &= 16\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 16.