Differentiation And Intergration In Determinants

Only the first row of $g(x)$ depends on $x$. Row-wise differentiation gives:

$$g^{\prime }(x)=\left|\begin{array}{ccc}{A}^{\prime }(x)& B^{\prime }(x)& C^{\prime }(x)\\ A(\alpha )& B(\alpha )& C(\alpha )\\ A^{\prime }(\alpha )& B^{\prime }(\alpha )& C^{\prime }(\alpha )\end{array}\right|$$

(The other two terms vanish because $R_2$ and $R_3$ are constant with respect to $x$.)

Now evaluate at $x=\alpha$:

• $g(\alpha )=0$: rows $R_1$ and $R_2$ become identical, so the determinant vanishes.
• $g^{\prime }(\alpha )=0$: rows $R_1$ and $R_3$ become identical, so the determinant vanishes.

Since $g(\alpha )=0$ and $g^{\prime }(\alpha )=0$, the polynomial $g(x)$ has $\alpha$ as a double root, i.e. $g(x)=(x-\alpha {)}^{2}P(x)$ where $P(x)$ is a polynomial of degree $3$.

Also, $f(x)=0$ is a quadratic with repeated root $\alpha$, so $f(x)=k(x-\alpha {)}^2$ for some constant $k$. Therefore $g(x)$ is divisible by $f(x)$.

Every row depends on $x$. Row-wise differentiation:

$$F^{\prime }(x)=\left|\begin{array}{ccc}{f}_1^{\prime }(x)& f_2^{\prime }(x)& f_3^{\prime }(x)\\ g_1(x)& g_2(x)& g_3(x)\\ h_1(x)& h_2(x)& h_3(x)\end{array}\right|+\left|\begin{array}{ccc}{f}_1(x)& f_2(x)& f_3(x)\\ g_1^{\prime }(x)& g_2^{\prime }(x)& g_3^{\prime }(x)\\ h_1(x)& h_2(x)& h_3(x)\end{array}\right|+\left|\begin{array}{ccc}{f}_1(x)& f_2(x)& f_3(x)\\ g_1(x)& g_2(x)& g_3(x)\\ h_1^{\prime }(x)& h_2^{\prime }(x)& h_3^{\prime }(x)\end{array}\right|$$

Now substitute $x=a$. Given that $f_r(a)=g_r(a)=h_r(a)$ for all $r$:

• In the first term, $R_2$ and $R_3$ both become $(f_1(a),f_2(a),f_3(a))$ - identical rows, so the first determinant is $0$.
• In the second term, $R_1$ and $R_3$ are identical, so it is $0$.
• In the third term, $R_1$ and $R_2$ are identical, so it is $0$.

Therefore $F^{\prime }(a)=0+0+0=0$.

Since multiple rows depend on $x$, we must expand first. Operate $R_1\to R_1-\sec x\cdot R_3$:

New $R_1$ becomes $(0,0,{\sec }^{2}x+\cot x\csc x-\cos x)$.

$$f(x)=\left|\begin{array}{ccc}0& 0& {\sec }^{2}x+\cot x\csc x-\cos x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$$

Expanding along $R_1$ (only the third entry is nonzero):

$$f(x)=({\sec }^{2}x+\cot x\csc x-\cos x)({\cos }^{4}x-{\cos }^{2}x)$$

Simplify ${\sec }^{2}x+\cot x\csc x-\cos x$: writing everything with $\sin$ and $\cos$,

$${\sec }^{2}x+\cot x\csc x-\cos x=\frac{1}{{\cos }^{2}x}+\frac{\cos x}{{\sin }^{2}x}-\cos x=\frac{{\sin }^{2}x+{\cos }^{3}x-{\cos }^{3}x{\sin }^{2}x}{{\cos }^{2}x{\sin }^{2}x}\cdot {\cos }^{2}x$$

After algebraic simplification (multiply out and factor ${\cos }^{2}x-1=-{\sin }^{2}x$):

$$f(x)=-({\sin }^{2}x+{\cos }^{5}x)$$

Now integrate from $0$ to $\pi /2$:

$${\int }_0^{\pi /2}f(x)dx=-{\int }_0^{\pi /2}({\sin }^{2}x+{\cos }^{5}x)dx$$

Using standard results ${\int }_0^{\pi /2}{\sin }^{2}xdx=\frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi }{4}$ and ${\int }_0^{\pi /2}{\cos }^{5}xdx=\frac{4\cdot 2}{5\cdot 3\cdot 1}=\frac{8}{15}$:

$${\int }_0^{\pi /2}f(x)dx=-\left(\frac{\pi }{4}+\frac{8}{15}\right)$$

Split $R_3$ using Property 5:

$$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ bc& ac& ab\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$$

Apply $C_1\to C_1-C_2$ and $C_3\to C_3-C_2$ to both determinants:

$$\Delta =\left|\begin{array}{ccc}0& 1& 0\\ a-b& b& c-b\\ bc-ca& ac& ab-ca\end{array}\right|+\left|\begin{array}{ccc}0& 1& 0\\ a-b& b& c-b\\ a^2-b^2& b^2& c^2-b^2\end{array}\right|$$

Factor $(a-b)$ from column 1 and $(c-b)$ from column 3 in both determinants:

$$\Delta =(a-b)(b-c)\left(-\left|\begin{array}{ccc}0& 1& 0\\ 1& b& -1\\ -c& ac& a\end{array}\right|-\left|\begin{array}{ccc}0& 1& 0\\ 1& b& -1\\ a+b& b^2& -(a+b)\end{array}\right|\right)$$

Expanding each along $R_1$ (only the middle entry contributes, with sign $-1$):

$$=(a-b)(b-c)[(c-a)+(c-a)]=2(a-b)(b-c)(c-a)$$

The determinant $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ is the standard circulant, which factors as:

$$\Delta =-\frac{1}{2}(a+b+c)\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]=0$$

So either $a+b+c=0$ or $a=b=c$.

Case A: $a+b+c=0$. Taking real and imaginary parts:

$$\cos \theta +\cos 2\theta +\cos 3\theta =0,\sin \theta -\sin 2\theta +\sin 3\theta =0$$

Grouping: $\cos \theta +\cos 3\theta =2\cos 2\theta \cos \theta$, so the cosine equation becomes $\cos 2\theta (2\cos \theta +1)=0$. Similarly the sine equation becomes $\sin 2\theta (1-2\cos \theta )=0$.

These two must hold simultaneously. $\cos 2\theta =0$ requires $\sin 2\theta \ne 0$ (else $2\theta$ is undefined mod $\pi$), and $\cos \theta =1/2$ contradicts $\cos \theta =-1/2$. So Case A has no solution.

Case B: $a=b=c$, i.e. $e^{i\theta }=e^{-2i\theta }=e^{3i\theta }$. This forces $e^{i\theta }=1$, giving $\cos \theta =1$ and $\sin \theta =0$, i.e. $\theta =2n\pi$, $n\in \mathbb{Z}$.