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Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\ -\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0 & 0 & 1 & a^2 & b^2 \\ 0 & 0 & 0 & a & b \\ 0 & 0 & 0 & -b & a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$

I have trouble solving the determinant.. But what is immediately visible are those zeroes in the matrix, just one more zero is needed such that this matrix is a triangular matrix (the element $a_{54}$ must be zero for this but it is $-b$ instead). If it was zero we could just multiply the diagonal and the product would be our determinant.

I have tried various ways to form this matrix such that $a_{54}$ is zero but the way I formed harmed the matrix and I got a wrong determinant as solution : /

As example, I have multiplied row $4$ with $b$, multiply row $5$ with $a$ and then add row $4$ to row $5$. Because I multiplied row $5$ with $a$, I need to divide the determinant by $a$ at the end.

So then I have the matrix $$\begin{pmatrix} \sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab\\ -\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2\\ 0 & 0 & 1 & a^2 & b^2\\ 0 & 0 & 0 & a & b\\ 0 & 0 & 0 & 0 & a^2+b^2 \end{pmatrix}$$

$$\text{Thus }\det = \frac{\sin\alpha \cdot \sin\alpha \cdot 1 \cdot a \cdot (a^2+b^2)}{a}=\sin^2\alpha \cdot (a^2+b^2)$$

But this is wrong and I don't see how to get the correct determinant...?

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    $\begingroup$ You've still got one element below the diagonal (the $-\cos{\alpha}$), which also needs to be expanded. $\endgroup$
    – Chappers
    Commented May 14, 2018 at 19:28

5 Answers 5

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Using determinant of block matrix

$$\det\begin{pmatrix}B&C\\ 0& D\end{pmatrix}=\det(B)\det(D)$$ we get that the desired determinant is

$$\det\begin{pmatrix}\sin\alpha&\cos\alpha\\ -\cos\alpha& \sin\alpha\end{pmatrix}\det(1)\det\begin{pmatrix}a&b\\ -b& a\end{pmatrix}=a^2+b^2$$

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same as determinant of $$W=\begin{pmatrix} \sin\alpha & \cos\alpha & 0 & ? & ?? \\ -\cos\alpha & \sin\alpha & 0 & ??? & ???? \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & a & b \\ 0 & 0 & 0 & -b & a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$

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Your matrix is block upper triangular so

$$\begin{vmatrix} \sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\ -\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0 & 0 & 1 & a^2 & b^2 \\ 0 & 0 & 0 & a & b \\ 0 & 0 & 0 & -b & a \end{vmatrix}= \underbrace{\begin{vmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{vmatrix}}_{=1}\cdot 1 \cdot \underbrace{\begin{vmatrix} a & b \\ -b & a \end{vmatrix}}_{=a^2+b^2} = a^2 + b^2$$

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$$\det \left[\begin{array}{cc|c|cc} \sin (\alpha) & \cos (\alpha) & a \,\sin (\alpha) & b\,\cos (\alpha) & a b \\ -\cos (\alpha) & \sin (\alpha) & - a^2 \sin (\alpha) & b^2 \cos (\alpha) & a^2 b^2\\ \hline 0 & 0 & 1 & a^2 & b^2\\ \hline 0 & 0 & 0 & a & b \\ 0 & 0 & 0 & -b & a\end{array}\right] = \\ = \underbrace{\det \begin{bmatrix} \sin (\alpha) & \cos (\alpha)\\ -\cos (\alpha) & \sin (\alpha)\end{bmatrix}}_{= 1} \cdot \det (1) \cdot \det \begin{bmatrix} a & b\\ -b & a\end{bmatrix} = a^2 + b^2$$

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If you calculate along the first column, \begin{align} |A|&= \sin\alpha \begin{vmatrix} \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0 & 1 & a^2 & b^2 \\ 0 & 0 & a & b \\ 0 & 0 & -b & a \end{vmatrix} +\cos\alpha\begin{vmatrix} \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\ 0 & 1 & a^2 & b^2 \\ 0 & 0 & a & b \\ 0 & 0 & -b & a \end{vmatrix} \\ \ \\ &=(\sin^2\alpha+\cos^2\alpha)\,\begin{vmatrix} 1 & a^2 & b^2 \\ 0 & a & b \\ 0 & -b & a \end{vmatrix} \\ \ \\ &=\begin{vmatrix} 1 & a^2 & b^2 \\ 0 & a & b \\ 0 & -b & a \end{vmatrix} \\ \ \\ &=a^2+b^2. \end{align}

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  • $\begingroup$ Yes, there should have been no fifth column. Thanks for noticing. $\endgroup$ Commented May 14, 2018 at 20:02

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