Question

Reduce the quadratic form to canonical form, \(3x_1^2+ 2x_2^2+8x_{12}+8x_{23}+8x_{31}=0\).

(a) \(\begin{bmatrix}3 & 4 & 4 \\4 & 0 & 4 \\ 4 & 4 & 2\end{bmatrix} \)

(b) \(\begin{bmatrix}3 & 4 & 4 \\4 & 2 & 0 \\ 4 & 4 & 0\end{bmatrix} \)

(c) \(\begin{bmatrix}3 & 4 & 4 \\4 & 2 & 4 \\ 4 & 4 & 0\end{bmatrix} \)

(d) \(\begin{bmatrix}3 & 4 & 0 \\4 & 2 & 4 \\ 4 & 4 & 0\end{bmatrix} \)

I got this question in an internship interview.

This key question is from Canonical Form or Sum of Squares Form in chapter Eigen Values and Eigen Vectors of Engineering Mathematics

Answer 1

Right choice is (c) \(\begin{bmatrix}3 & 4 & 4 \\4 & 2 & 4 \\ 4 & 4 & 0\end{bmatrix} \)

For explanation I would say: Given quadratic form is, \(3x_1^2+ 2x_2^2+8x_{12}+8x_{23}+8x_{31}=0\)

General form of the matrix can be written as, \(\begin{bmatrix}x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33}\end{bmatrix} \)  

Hence, the matrix form can be obtained by,

Placing the square term coefficients in the diagonal of the matrix such that, \(x_{ii}= x_i^2,\)

\(\begin{bmatrix}3 & x_{12} & x_{13} \\ x_{21} & 2 & x_{23} \\ x_{31} & x_{32} & 0\end{bmatrix} \)  

Dividing the coefficients of terms xij between xij and xji positions, for example, the coefficient of x12 is 8, hence the term x12 =x21= 8/2 = 4.

Therefore, the matrix form of the given quadratic equation is,

\(\begin{bmatrix}3 & 4 & 4 \\ 4 & 2 & 4 \\ 4 & 4 & 0\end{bmatrix} \)