Question

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\).

a. \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

b. \(2{x_1}{x_2}\)

Short answer

1. The matrix for the quadratic form \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\) is \(\left( {\begin{aligned}{{}}5&8\\8&{ - 5}\end{aligned}} \right)\).

2. The matrix for the quadratic form \(2{x_1}{x_2}\) is \(\left( {\begin{aligned}{{}}0&1\\1&0\end{aligned}} \right)\).

Step-by-step solution

Matrix of the quadratic form

The coefficients of the square of variables, that is \(x_i^2\), are to be divided on the main diagonal as per the order, and the coefficient of term \({x_i}{x_j}{\rm{ }}\left( {i \ne j} \right)\), divided by 2 to split properly between the entries \(\left( {i,j} \right)\) and \(\left( {j,i} \right)\).

Find the corresponding matrix of \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

(a)

The given quadratic expression is \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\).

For this expression, the order of the square matrix is 2.

So, let \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{aligned}} \right)\)

According to the rule in steps (1), 5 and \( - 5\), will be on the main diagonal of the matrix, like \({a_{11}} = 5\) and \({a_{22}} = - 5\).

And \({a_{12}} = \frac{{16}}{2}\), \({a_{21}} = \frac{{16}}{2}\) .

Substitute the required value into \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{aligned}} \right)\).

\(A = \left( {\begin{aligned}{{}}5&8\\8&{ - 5}\end{aligned}} \right)\)

So, the required matrix is \(\left( {\begin{aligned}{{}}5&8\\8&{ - 5}\end{aligned}} \right)\).

Find the corresponding matrix of \(2{x_1}{x_2}\)

(b)

The given quadratic expression is \(2{x_1}{x_2}\), which can be written as \(0x_1^2 + 2{x_1}{x_2} + 0x_2^2\).

For this expression, the order of the square matrix is 2.

So, let \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{aligned}} \right)\)

According to the rule in steps (1), 3 and 0 will be on the main diagonal of the matrix, like \({a_{11}} = 0\) and \({a_{22}} = 0\).

And \({a_{12}} = \frac{2}{2}\), \({a_{21}} = \frac{2}{2}\) .

Substitute the required value into \(A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{aligned}} \right)\).

\(A = \left( {\begin{aligned}{{}}0&1\\1&0\end{aligned}} \right)\)

So, the required matrix is \(\left( {\begin{aligned}{{}}0&1\\1&0\end{aligned}} \right)\).