Question

Compute the quadratic form \({x^T}Ax\), when \(A = \left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\) and

a. \(x = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

b. \(x = \left( {\begin{aligned}{{}}{ - 2}\\{ - 1}\\5\end{aligned}} \right)\)

c. \(x = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\)

Short answer

a. For \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 3x_1^2 + 2x_2^2 + 4{x_1}{x_2} + 2{x_2}{x_3}\)

b.For \({\bf{x}} = \left( {\begin{aligned}{{}}{ - 2}\\{ - 1}\\5\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = 12\)

c.For \({\bf{x}} = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\): \({{\bf{x}}^T}A{\bf{x}} = \frac{{11}}{2}\)

Step-by-step solution

Find \({x^T}Ax\)

The given matrix is\(A = \left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\), and the given vector is \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\).

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

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= {\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)^T}\left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}&{{x_3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}&{{x_3}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{3{x_1} + 2{x_2}}\\{2{x_1} + 2{x_2} + {x_3}}\\{{x_2}}\end{aligned}} \right)\\ &= {x_1}\left( {3{x_1} + 2{x_2}} \right) + {x_2}\left( {2{x_1} + 2{x_2} + {x_3}} \right){x_3}{x_2}\\ &= 3x_1^2 + 2x_2^2 + 4{x_1}{x_2} + 2{x_2}{x_3}\end{aligned}\)

Hence, the expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(3x_1^2 + 2x_2^2 + 4{x_1}{x_2} + 2{x_2}{x_3}\).

Find \({x^T}Ax\) for \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

(b).

As expression for \({{\bf{x}}^T}A{\bf{x}}\) is \(3x_1^2 + 2x_2^2 + 4{x_1}{x_2} + 2{x_2}{x_3}\) when \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\).

So\({\bf{x}} = \left( {\begin{aligned}{{}}{ - 2}\\{ - 1}\\5\end{aligned}} \right)\), simplify \(3x_1^2 + 2x_2^2 + 4{x_1}{x_2} + 2{x_2}{x_3}\) by substituting \({x_1} = - 2,{x_2} = - 1,{x_3} = 5\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 3{\left( { - 2} \right)^2} + 2{\left( { - 1} \right)^2} + 4\left( { - 2} \right)\left( { - 1} \right) + 2\left( { - 1} \right)\left( 5 \right)\\ &= 12\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is 12.

Find \({x^T}Ax\) for  \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

(c)

For \({\bf{x}} = \left( {\begin{aligned}{{}}{{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right.} {\sqrt 2 }}}\\{{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right.} {\sqrt 2 }}}\\{{1 \mathord{\left/ {\vphantom {1 {\sqrt 2 }}} \right.} {\sqrt 2 }}}\end{aligned}} \right)\), simplify \(3x_1^2 + 2x_2^2 + 4{x_1}{x_2} + 2{x_2}{x_3}\) by substituting \({x_1} = \frac{1}{{\sqrt 2 }},{x_2} = \frac{1}{{\sqrt 2 }},{x_3} = \frac{1}{{\sqrt 2 }}\).

\(\begin{aligned}{}{{\bf{x}}^T}A{\bf{x}} &= 3{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 4\left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right) + 2\left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)\\ &= \frac{{11}}{2}\end{aligned}\)

So, the value of \({{\bf{x}}^T}A{\bf{x}}\) is \(\frac{{11}}{2}\).