Answer to Question #91094 in Linear Algebra for Kamlesh Devi

Question

Reduce the following equation to standard form hence identify the conic it represents x2 - 3xy + y2 + 4x - 4y - 5 = 0

Answer 1

equation:

$$x^2-3xy+y^2+4x-4y-5=0$$

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

Find eigenvalues of matrix

$$A_{33}=\begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}$$

from equation

$$\begin{vmatrix} A-z & B/2 \\ B/2 & C-z \end{vmatrix}=0$$

substituting A = 1, B = -3, C = 1 in this equation

$$\begin{vmatrix} 1-z & -3/2 \\ -3/2 & 1-z \end{vmatrix}=0$$

$$(1-z)^2-(-3/2)^2=0$$

$$(1-3/2-z)(1+3/2-z)=0$$

$$(z+1/2)(z-5/2)=0$$

$$z_1=-1/2, z_2=5/2$$

Let

$$A_q=\begin{pmatrix} A& B/2 &D/2\\ B/2 & C&E/2\\ D/2&E/2&F \end{pmatrix}= \begin{pmatrix} 1& -3/2 &2\\ -3/2 & 1&-2\\ 2&-2&-5 \end{pmatrix}$$

Standard form of equation

$$x^2-3xy+y^2+4x-4y-5=0$$

can be found using formula:

$$z_1x'^2+z_2y'^2=-detA_q/detA_{33}$$

$$detA_{33}=\begin{vmatrix} 1& -3/2 \\ -3/2 & 1 \end{vmatrix}=1-9/4=-5/4$$

$$detA_q=\begin{vmatrix} 1& -3/2 &2\\ -3/2 & 1&-2\\ 2&-2&-5 \end{vmatrix}=$$

add the first row to the second

$$\begin{vmatrix} 1& -3/2 &2\\ -1/2 & -1/2&0\\ 2&-2&-5 \end{vmatrix}=$$

subtract the first column from the second

$$\begin{vmatrix} 1& -5/2 &2\\ -1/2 & 0&0\\ 2&-4&-5 \end{vmatrix}=$$

$$-(-1/2)(-5/2*(-5)-(-4)*2)=41/4$$

Substitute all found values in the equation

$$z_1x'^2+z_2y'^2=-detA_q/detA_{33}$$

$$-1/2x'^2+5/2y'^2=-(41/4)/(-5/4)=41/5$$

dividing by (41/5)

$$-1/2/(41/5)x'^2+5/2/(41/5)y'^2=1$$

$$-x'^2/(82/5)+y'^2/(82/25)=1$$

since coefficients of squared terms have different signs, this is standard equation of hyperbola.

Answer: standard equation:

$$-x'^2/(82/5)+y'^2/(82/25)=1$$

conic section is hyperbola.