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Found in: Page 395

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Find the matrix of the quadratic form. Assume x is in $${\mathbb{R}^2}$$.a. $$3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}$$ b. $$6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}$$

1. The matrix for the quadratic form $$3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}$$ is \left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right).

1. The matrix for the quadratic form $$6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}$$ is \left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right).
See the step by step solution

## Step 1: 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 $$\left( {j,i} \right)$$.

## Step 2: Find the corresponding matrix of $$5x_1^2 + 16{x_1}{x_2} - 5x_2^2$$

(a)

The given quadratic expression is $$3x_1^2 + 2x_2^2 - 5x_3^2 - 6{x_1}{x_2} + 8{x_1}{x_3} - 4{x_2}{x_3}$$.

For this expression, the order of the square matrix is 3.

So, let A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)

According to the rule in steps (1), 3, 2, and $$- 5$$, will be on the main diagonal of the matrix, like $${a_{11}} = 3$$, $${a_{22}} = 2$$ and $${a_{33}} = - 5$$.

And,$${a_{12}} = \frac{{ - 6}}{2}$$, $${a_{21}} = \frac{{ - 6}}{2}$$, $${a_{13}} = \frac{8}{2}$$, $${a_{31}} = \frac{8}{2}$$ and $${a_{23}} = \frac{{ - 4}}{2}$$, $${a_{32}} = \frac{{ - 4}}{2}$$.

Substitute the required value into A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right).

A = \left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right)

So, the required matrix is \left( {\begin{aligned}{{}}3&{ - 3}&4\\{ - 3}&2&{ - 2}\\4&{ - 2}&{ - 5}\end{aligned}} \right).

## Step 3: Find the corresponding matrix of $$2{x_1}{x_2}$$

(b)

The given quadratic expression is $$6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}$$, which can be written as $$0x_1^2 + 0x_2^2 + 0x_3^2 + 6{x_1}{x_2} + 4{x_1}{x_3} - 10{x_2}{x_3}$$.

For this expression, the order of the square matrix is 3.

So, let A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right)

According to the rule in step (1), 0, 0 and 0, will be on the main diagonal of the matrix, like $${a_{11}} = 0$$, $${a_{22}} = 0$$ and $${a_{33}} = 0$$.

And $${a_{12}} = \frac{6}{2}$$, $${a_{21}} = \frac{6}{2}$$, $${a_{13}} = \frac{2}{2}$$, $${a_{31}} = \frac{2}{2}$$ and $${a_{23}} = \frac{{ - 10}}{2}$$, $${a_{32}} = \frac{{ - 10}}{2}$$ .

Substitute the required value into A = \left( {\begin{aligned}{{}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{aligned}} \right).

A = \left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right)

So, the required matrix is \left( {\begin{aligned}{{}}0&3&2\\{ - 3}&0&{ - 5}\\2&{ - 5}&0\end{aligned}} \right).