# Show that the determinant is equal to 0

• MHB
• mathmari
In summary, Wolfram concludes that $\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$ has a geometric extremum at $\cos\alpha=\cos\beta=\cos\gamma$.
mathmari
Gold Member
MHB
Hey!

Let $\alpha, \beta, \gamma$ be internal angles of an arbitrary triangle.

I want to show that $$\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$$

We have the following:
\begin{align*}&\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=\cos \beta \cdot \det \begin{pmatrix} -1 & \cos\alpha \\ \cos\gamma & \cos\beta\end{pmatrix}-\cos \alpha \cdot \det \begin{pmatrix} \cos\gamma & \cos\alpha \\ -1 & \cos\beta\end{pmatrix}+(-1)\cdot \det \begin{pmatrix} \cos\gamma & -1 \\ -1 & \cos\gamma \end{pmatrix} \\ & = \cos \beta \cdot \left (-\cos \beta-\cos\gamma\cdot \cos\alpha\right )-\cos \alpha \cdot \left (\cos\gamma\cdot \cos\beta-(-1)\cdot \cos\alpha\right )-\left (\cos^2 \gamma -(-1)^2\right ) \\ & =-\cos^2 \beta-\cos\gamma\cdot \cos\alpha\cdot \cos \beta-\cos\alpha\cdot \cos\gamma\cdot \cos\beta- \cos^2\alpha-\cos^2 \gamma +1 \\ & =1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma \end{align*}

Do we use here the Law of cosines?
\begin{align*}&c^2=a^2+b^2-2ab\cdot \cos \gamma \Rightarrow \cos \gamma=\frac{a^2+b^2-c^2}{2ab} \\ &b^2=a^2+c^2-2ac\cdot \cos \beta \Rightarrow \cos \beta=\frac{a^2+c^2-b^2}{2ac} \\ &a^2=b^2+c^2-2bc\cdot \cos \alpha \Rightarrow \cos\alpha=\frac{b^2+c^2-a^2}{2bc}\end{align*}

Or how could we continue? (Wondering)

Hey mathmari! (Smile)

That should work I think although it looks like the expression will get pretty complicated first.
Alternatively we might try to use $α+β+γ=\pi$.
That is, substitute $γ=\pi-α-β$. (Thinking)

I like Serena said:
That should work I think although it looks like the expression will get pretty complicated first.
Alternatively we might try to use $α+β+γ=\pi$.
That is, substitute $γ=\pi-α-β$. (Thinking)

Do you mean $\cos \gamma=\cos (\pi -\alpha-\beta)=-\cos (\alpha+\beta )$ ? (Wondering)

mathmari said:
Do you mean $\cos \gamma=\cos (\pi -\alpha-\beta)=-\cos (\alpha+\beta )$ ?

That is what I was thinking yes.
I didn't try to calculate it manually though. It's just an alternative approach.
However, I did find that it is sufficient for Wolfram to conclude that it's zero, so it seems likely that filling it in by hand is feasible. (Thinking)

Another alternative approach is to try to find and find a geometric reason why the vectors are linearly dependent.
I didn't find one yet though. (Thinking)

mathmari said:
$$1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma$$

Doesn't this expression have cyclic symmetry? Implying there is an extremum at $\cos\alpha=\cos\beta=\cos\gamma$? Then, applying La Grange multipliers to the general case $\alpha+\beta+\gamma=\pi$, we arrive at the desired result.

## 1. What is a determinant?

A determinant is a mathematical concept used to measure the size and orientation of a matrix. It is denoted by det(A) or |A| and is calculated using a specific formula.

## 2. Why is it important to show that the determinant is equal to 0?

Showing that the determinant is equal to 0 can help determine if a matrix is singular or not. A determinant of 0 indicates that the matrix is not invertible, which can have important implications in solving equations and performing other mathematical operations.

## 3. How do you prove that the determinant is equal to 0?

To prove that the determinant is equal to 0, you can use various methods such as expanding the determinant using cofactor expansion or row reduction. These methods involve manipulating the elements of the matrix to simplify the calculation and eventually show that the determinant is equal to 0.

## 4. Can the determinant be equal to 0 for any size of matrix?

Yes, the determinant can be equal to 0 for any size of matrix. However, for a square matrix, the determinant is only equal to 0 if the matrix is singular. For non-square matrices, the determinant is defined differently and can also be equal to 0.

## 5. How is the determinant related to the properties of a matrix?

The determinant is related to several properties of a matrix, such as its invertibility, rank, and eigenvalues. It can also be used to solve systems of linear equations and determine whether a set of vectors is linearly independent or not.

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