Needs some help on this

Mathematics

2 answers:

kari74 [83]3 years ago

8 0

A is the answer to the question.

san4es73 [151]3 years ago

3 0

He’s wrong “A” is the answe because I took the same test man

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consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cros

Bezzdna [24]

Answer:

$q(x,y,z)=16x^{2}-5y^{2}-5z^{2}$

Step-by-step explanation:

The given quadratic form is of the form

$q(x,y,z)=ax^2+by^2+dxy+exz+fyz$ .

Where $a=11,b=-1,c=-4,d=-16,e=8,f=-4$ .Every quadratic form of this kind can be written as

$q(x,y,z)={\bf x}^{T}A{\bf x}=ax^2+by^2+cz^2+dxy+exz+fyz=\left(\begin{array}{ccc}x&y&z\end{array}\right) \left(\begin{array}{ccc}a&\frac{1}{2} d&\frac{1}{2} e\\\frac{1}{2} d&b&\frac{1}{2} f\\\frac{1}{2} e&\frac{1}{2} f&c\end{array}\right) \left(\begin{array}{c}x&y&z\end{array}\right)$

Observe that $A$ is a symmetric matrix. So $A$ is orthogonally diagonalizable, that is to say, $D=Q^{T}AQ$ where $Q$ is an orthogonal matrix and $D$ is a diagonal matrix.

In our case we have:

$A=\left(\begin{array}{ccc}11&(\frac{1}{2})(-16) &(\frac{1}{2}) (8)\\(\frac{1}{2}) (-16)&(-1)&(\frac{1}{2}) (-4)\\(\frac{1}{2}) (8)&(\frac{1}{2}) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)$

The eigenvalues of $A$ are $\lambda_{1}=16,\lambda_{2}=-5,\lambda_{3}=-5$ .

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

$Q=\left(\begin{array}{ccc}\frac{4}{\sqrt{21}}&-\frac{1}{\sqrt{17}}&\frac{8}{\sqrt{357}}\\\frac{-2}{\sqrt{21}}&0&\sqrt{\frac{17}{21}}\\\frac{1}{\sqrt{21}}&\frac{4}{\sqrt{17}}&\frac{2}{\sqrt{357}}\end{array}\right)$

$D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)$

Now, we have to do the change of variables ${\bf x}=Q{\bf y}$ to obtain

$q({\bf x})={\bf x}^{T}A{\bf x}=(Q{\bf y})^{T}AQ{\bf y}={\bf y}^{T}Q^{T}AQ{\bf y}={\bf y}^{T}D{\bf y}=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\lambda_{3}y_{3}^{2}=16y_{1}^{2}-5y_{2}^{2}-5y_{3}^2$